**Area between two curves: Based on formula given in Application of Integrals**

**CASE – 1:**

For finding the **Area** bounded by the curve **y = f(x),** **x-axis** and the lines **x = a**, **x = b** let us consider a very thin vertical strip of length **y** and width **dx**. Therefore, **Area** of the strip **(dA) = y dx [Since, y = f(x)]**

Hence the total **Area** enclosed by the curve **y = f(x), x- axis** and the lines **x = a, x = b**:

**\(\boldsymbol{A =\int_{a}^{b} dA=\int_{a}^{b}y\;dx}\)**

**Therefore, A = \(\int_{a}^{b}f(x)\;dx\)**

** **

**CASE – 2:**

** **

Similarly the **Area** bounded by the curve **x = g(y), y-axis** and the **lines y = c, y = d **is given by:

**\(A=\int_{c}^{d} dA=\int_{c}^{d} x\;dy\)**

**Therefore, A = \(\int_{c}^{d}g(y)\;dy\)**

** **

**CASE – 3:**

** **

If the curve lies below **x-axis**, then the **Area** bounded by the curve **y = f(x)**, **x-axis** and the lines **x = a**, **x = b** will come negative.

Since, the **area** cannot be negative therefore we will neglect the negative sign and considering its **absolute value** only.

**\(A = \left |\int_{a}^{b}f(x)\;dx \right|\)**

** **

**CASE – 4:**

As shown above, some portion of the curve lies above the x-axis and some portion of the curve lies below the x-axis.

Here in this case, **Area-1 > 0** and **Area-2 < 0**

Therefore, **total** Area bounded by the **curve y= f(x)**, **x-axis** and the lines **x=a**, **x=b** is given by: **A = |A|+B**

** **

**NOTE:**

**\(\boldsymbol{\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}}\)**

** **

**Example 1: Find the area enclosed by equation: x ^{2 }+ y^{2} = 9**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **x ^{2 }+ y^{2} = 3^{2}** represents the

**Therefore, y = \(\sqrt{3^{2}-x^{2}}\)**

** **

**From the figure, the Area enclosed by the circle = 4 ****×**** (Area enclosed by the curve ABOA)**

**Now, Area enclosed by the curve** **ABOA**:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx = \int_{0}^{3} \sqrt{9-x^{2}}\;dx\)

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin^{-1}\frac{x}{3} \right | _{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin^{-1}(1) – \frac{0}{2} \sqrt{9-0}-\frac{0}{2} \sin^{-1}\frac{0}{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9}{2}\times \frac{\pi }{2}=\frac{9\pi }{4}}\) **unit ^{2}**

**Therefore, the Area enclosed by the circle =** **4 ****×**** (area enclosed by curve ABOA)**

**\(\boldsymbol{\Rightarrow }\) \(4\times \frac{9\pi }{4}=9\pi\)**

**Hence the Area enclosed by the circle = 9****π**** unit ^{2}**

**Example 2: Find the area enclosed by the curve \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)**

** **

**Sol:Based on formula given in Application of Integrals**

**Equation** **\(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)** **represents an** **ellipse with major axis = 3 units and minor axis = 2 units**

**Since, \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)**

**\(\boldsymbol{\Rightarrow }\) 4x ^{2} + 9y^{2} = 36**

**\(\boldsymbol{\Rightarrow }\) 9y ^{2} = 6^{2 }– (2x)^{2}**

**Therefore, \(y=\sqrt{\frac{6^{2}-(2x)^{2}}{9}}\)**

**Therefore, the Area of region enclosed by the ellipse = 4 ****× ****(Area enclosed by the curve ABOA)**

Now, **Area enclosed by the curve ABOA = \(\int_{0}^{3} y\;dx\)**

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\times \int_{0}^{3}\sqrt{6^{2}-(2x)^{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\int_{0}^{3}\;\frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\left | \frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6} \right |_{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}[\frac{6}{2}\sqrt{36-36}+18\sin^{-1}(1)]=3\pi }\)

**Therefore, the Area enclosed by the** **curve ABOA = 3****π**** unit ^{2}**

**Hence, the total Area enclosed by an ellipse ABA’B’ = ****4****×****3****π** **= 12****π unit ^{2}.**

** **

**Area of region bounded by the curve and line:**

** **

**Example 3: Find the area of the region enclosed by the curve y = x ^{2} and the line y = 3.**

** **

**Sol: Based on formula given in Application of Integrals**

** **

**y = x ^{2} **represents a

i.e. **x ^{2} = 9**

Therefore, **x = +3 or -3**

**Hence the coordinates of point M = (3, 9)**

The area of the region bounded by the curve **y = x ^{2}** and the line

Now, **Area of region POMP = 2(area of region AOMA)**

Since, **x ^{2} = y**

Therefore, **x = \(\sqrt{y}\)**

Thus, **the Area of region bounded by the curve AOMA:**

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{9}x\;dy\) = \(\int_{0}^{9}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{0}^{9}=\frac{2}{3}\times (9^{\frac{3}{2}})}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times 27 = 18}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve AOMA = 18 unit ^{2}**

**Hence, the Area of region bounded by the curve POMP ****= 2 ****×**** (area of region bounded by curve AOMA) ****= 36 unit ^{2}**

** **

** **

**Example 4: Find the area enclosed by the curve x ^{2 }+ y^{2 }= 50, x-axis and the line y = x in the 1^{st} quadrant.**

** **

**Sol: Based on formula given in Application of Integrals**

Equation **x ^{2 }+ y^{2 }= 50** represents a

Since y = x

Therefore x^{2 }+ x^{2 }= 50 [for points of intersection of both the curves]

Hence, **x = +5 and -5**

Similarly, ** y = +5 and -5**

**Thus the coordinates of point B are (5, 5)**

** **

Form the figure, the area of region bounded by the curve **x ^{2 }+ y^{2 }= 50**,

**i.e. Area of triangle OMB + Area under the curve MBAM.**

**Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)**

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;\;=\;\frac{1}{2}\times 5\times 5 =\frac{25}{2}\) unit^{2}.

Now, the Area under the curve **MBAM** = \(\int_{5}^{5\sqrt{2}}y\;dx\)

Since, **x ^{2} +y^{2} = 50**. Therefore,

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{50-x^{2}}}\)

Therefore, the Area under curve **MBAM [x ^{2 }+ y^{2 }= 50]**:

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{5\sqrt{2}}\sqrt{50-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{50-x^{2}}+\frac{50}{2}\;\sin^{-1}\frac{x}{5\sqrt{2}} \right |_{5}^{5\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{5\sqrt{2}}{2}\times\ \sqrt{50-50}]+[\frac{50}{2}\;\sin^{-1}(1)]-[\frac{5}{2}\times \sqrt{50-25}]-[\frac{50}{2}\times \sin^{-1}\frac{1}{\sqrt{2}}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{0+\frac{25\pi }{2}-\frac{25}{2}-\frac{25\pi }{4}=\frac{25}{4}(\pi -2)}\)**unit ^{2}**

Therefore, the **Area under the curve MBAM = \(\boldsymbol{\frac{25}{4}(\pi -2)}\) unit ^{2}**

**Now, the total Area under the shaded region** **=** **Area of triangle OMB + Area under the curve MBAM**

\(\boldsymbol{\Rightarrow }\) \(\frac{25}{2}+\frac{25}{4}(\pi -2)=\frac{25}{2}+\frac{25\pi }{4}-\frac{25}{2}\boldsymbol{=\frac{25\pi }{4}}\)**unit ^{2}**

**Hence, the Area of shaded region OMABO = \(\boldsymbol{\frac{25\pi }{4}}\)unit ^{2}**

** **

** **

**Q.1: Find the area enclosed by the curve y = x ^{2} and the lines y = 2, y = 4 and the y-axis.**

** **

**Sol: Based on formula given in Application of Integrals**

Equation **y = x ^{2}** represents a

** **

The Area of the region bounded by the curve **y = x ^{2}**,

Now, the **Area of region AA’B’BA = 2 (Area of region ABNMA)**

Since, x^{2} = y

Therefore, **x = \(\sqrt{y}\)**

Thus, **the Area of region bounded by the curve ABNMA [****y = x ^{2}**

\(\boldsymbol{\Rightarrow }\) \(\int_{2}^{4}x\;dy\) = \(\int_{2}^{4}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{2}^{4}=\frac{2}{3}\times [(4^{\frac{3}{2}})-(2)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times (8-2\sqrt{2}) = \frac{4}{3}(4-\sqrt{2})}\) **unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABNMA = \(\boldsymbol{\frac{4}{3}(4-\sqrt{2})}\)unit ^{2}**

**Hence, the ****Area of region bounded by the curve AA’B’BA ****=**** 2(Area of region bounded by the curve ABNMA)****= \(\boldsymbol{\frac{8}{3}(4-\sqrt{2})}\)unit ^{2}**

** **

**Q.2: Find the area enclosed by the curve y ^{2} = 4x and lines x = 1, x = 3 and the x- axis in the first quadrant.**

** **

**Sol : Based on formula given in Application of Integrals**

** **

**Equation y ^{2} = 4x **represents a

The area of the region bounded by the curve **y ^{2} = 4x**,

Now, the **Area of region bounded by the curve ABCDA:**

Since, **y ^{2} = 4x**

Therefore, **y = \(2\sqrt{x}\)**

Hence, **the area of region bounded by the curve ABCDA [y ^{2} = 4x]:**

\(\boldsymbol{\Rightarrow }\) \(\int_{1}^{3}y\;dx\) = \(\int_{1}^{3}2\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=\frac{4}{3}\times [(3^{\frac{3}{2}})-(1)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4}{3}\times (3\sqrt{3}-1) = \frac{-4\;+\;12\sqrt{3}}{3}}\) **unit ^{2}**

**Therefore, the ****area**** of region ****bounded ****by the ****curve ABCDA \(\boldsymbol{= \frac{-4\;+\;12\sqrt{3}}{3}}\)unit ^{2}**

** **

** **

**Q.3: Find the value of k if the line x = k divides the area enclosed by the curve y ^{2} = 9x and the line x = 4 in to two equal parts.**

** **

**Sol: Based on formula given in Application of Integrals**

**Equation y ^{2} = 9x **represents a

Since, the line **x = k** divides the Area OCBB’O in to two **equal halves** and the curve is symmetrical to **x-axis**. Therefore**, Area of the region OADO = Area of the region ABCDA.**

The Area of the region bounded by the curve **y ^{2} = 9x** and the line

Since, **y ^{2} = 9x**

Therefore, **y = \(3\sqrt{x}\)**

Hence, **the Area of region bounded by the curve OADO:**

\(\boldsymbol{\Rightarrow }\) **\(\int_{0}^{k}y\;dx\) = \(\int_{0}^{k}3\sqrt{x}\;dx\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=2\times [(k^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2k^{\frac{3}{2}}}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve OADO = \(\boldsymbol{2k^{\frac{3}{2}}}\)unit ^{2}**

The Area of the region bounded by the curve **y ^{2} = 9x** and the lines

**Now, the Area of region bounded by the curve ABCDA [****y ^{2} = 9x**

\(\boldsymbol{\Rightarrow }\) ** \(\int_{k}^{4}y\;dx\) = \(\int_{k}^{4}3\sqrt{x}\;dx\)**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{k}^{4}=2\times [(4^{\frac{3}{2}})-(k)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})}\) **unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABCDA = ****\(\boldsymbol{\Rightarrow (16-2k^{\frac{3}{2}})}\) unit ^{2}**

**Since, the Area of region OADO = Area of region ABCDA [GIVEN]**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})=2k^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k^{\frac{3}{2}}=4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k = 4^{\frac{2}{3}}}\)

**Therefore, the value of k = \(\boldsymbol{4^{\frac{2}{3}}}\)**

** **

** **

**Q.4: Find the area enclosed by the curve y ^{2} = 16x, y-axis and the line y = 2.**

** **

**Sol: Based on formula given in Application of Integrals**

**Equation y ^{2} = 16x **represents a

The Area of the region bounded by the curve **y ^{2} = 16x**,

Now, the **Area of region **enclosed by the curve **OABO:**

Since, **y ^{2} = 16x**

Therefore, **x = \(\frac{y^{2}}{16}\)**

Thus, **the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{2}x\;dy\) = \(\int_{0}^{2}\frac{y^{2}}{16}\;dy\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{1}{16}\times \frac{y^{3}}{3}\right |_{0}^{2}=\frac{1}{16}\times [\frac{8}{3}-0}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}}\) **unit ^{2}**

**Therefore, ****Area**** of region ****bounded ****by the ****curve OABO \(\boldsymbol{=\frac{1}{6}}\)unit ^{2}**

** **

** **

**Q.5: Find the area enclosed by the curve y ^{2} = 25x and the line x = 3**

** **

**Sol: Based on formula given in Application of Integrals**

** **

**Equation y ^{2} = 25x **represents a

The area of the region bounded by the curve **y ^{2} = 25x** and

Now, the** Area of region BOCAB = 2(Area of region OABO)**

Since, y^{2} = 25x

Therefore, **y = \(5\sqrt{x}\)**

Hence, **the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx\) = \(\int_{0}^{3}5\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 5\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{0}^{3}=\frac{10}{3}\times [(3^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{10}{3}\times (3\sqrt{3})=10\sqrt{3}}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve OABO\(\boldsymbol{=10\sqrt{3}}\) unit ^{2}**

**Now, the Area of region BOCAB = ****2 ****×**** (Area of region OABO) ****\(\boldsymbol{=20\sqrt{3}}\) unit ^{2}**

** **

** **

**Q.6: Find the area enclosed by the curve x ^{2 }= 8y, y = 1, y = 9 and the y-axis in the first quadrant.**

** **

**Sol: Based on formula given in Application of Integrals**

** **

**Equation x ^{2} = 8y **represents a

The area of the region bounded by the curve **x ^{2} = 8y**,

Now, the **Area of the region ABDCA:**

Since, **x ^{2} = 8y**

Therefore, **x = \(2\sqrt{2y}\)**

Hence, **the Area of region bounded by the curve ABDCA:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{1}^{9}x\;dy\) = \(\int_{1}^{9}2\sqrt{2y}\;dy\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\sqrt{2}\times \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{9}=\frac{4\sqrt{2}}{3}\times [(9^{\frac{3}{2}})-(1)}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4\sqrt{2}}{3}\times (27-1)=104\times \frac{\sqrt{2}}{3}}\) **unit ^{2}**

Therefore, **Area** of region **bounded** by the **curve ABDCA ****\(\boldsymbol{=104\times \frac{\sqrt{2}}{3}}\)unit ^{2}**

** **

** **

**Q.7: Find the area bounded by the curve whose equation is x ^{2} = 9y and the line x = 6y – 3.**

** **

**Sol: Based on formula given in Application of Integrals**

** **

**Equation x ^{2} = 9y **represents a

The Area of the region bounded by parabola **x ^{2} = 9y** and the line

Since, the **parabola x ^{2} = 9y** and the

(6y-3)^{2} = 9y

**\(\boldsymbol{\Rightarrow }\) 36y ^{2} – 45y + 9 = 0**

By **Hit and Trial method** solutions of this quadratic equation are:

**y = 1 and y = \(\frac{1}{4}\)**

Hence, the **co-ordinates** of **point A** and **point C** are** (3,1)** and** \((\frac{-3}{2},\frac{1}{4})\)** respectively

Since, **x ^{2} = 9y**

Therefore, **x = \(3\sqrt{y}\)**

The **Area** of region bounded by the curve **ABCOA** = [**Area** of region **OBCbO** – **Area** of region **OCbO**] + [**Area** of region **ABOaA** – **Area** of region **OAaO**]

**The Area enclosed by the curve OBCbO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3} y\;dx = \boldsymbol{\int_{0}^{3}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(\frac{9}{2}+9)=\frac{9}{4}}\) **unit ^{2}**

**The Area enclosed by the curve ABOaA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx} = \boldsymbol{\int_{\frac{-3}{2}}^{0}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(0-\frac{9}{8}+\frac{9}{2})=\frac{9}{16}}\) **unit ^{2}**

**The Area ****enclosed by the curve ****OAaO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx\; =\;\int_{\frac{-3}{2}}^{0} \frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{27}(0-\frac{-27}{8})=\frac{1}{8}}\)**unit ^{2}**

**The Area ****enclosed by the curve ****OCbO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3} y\;dx\; = \;\int_{0}^{3}\frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}(\frac{27}{3}-0)=1}\)** unit ^{2}**

**Since,** **the Area of region bounded by the curve ABCOA = [Area of region OBCbO – Area of region OCbO] + [Area of region ABOaA – Area of region OAaO]**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{9}{4}-1]+[\frac{9}{16}-\frac{1}{8}]=\frac{27}{16}}\)

**Therefore, the Area of region bounded by the curve ABCOA \(=\frac{27}{16}\) unit ^{2}**

**Q.8: Find the area enclosed by the curve 4y = x ^{2} and y = |x|**

** **

**Sol:Based on formula given in Application of Integrals**

** **

**Equation x ^{2} = 4y **represents a

The area of the region bounded by the curve **x ^{2} = 4y** and

Now, **Area of region OAEO = OABO – OEABO**

Since, **x ^{2} = 4y**

\(\boldsymbol{\Rightarrow }\) \(y=\frac{x^{2}}{4}\)

Now, **the Area of region bounded by the curve OEABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;=\;\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{4}\left | \frac{x^{3}}{3} \right |_{0}^{4}=\frac{16}{3}}\) **unit ^{2}**

**Therefore, the area of region bounded by the curve OEABO = \(=\frac{16}{3}\) unit ^{2}**

**Now, the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;\Rightarrow \;\int_{0}^{4}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{4}=8}\) **unit ^{2}**

**Therefore, the Area of the region bounded by the curve OABO = 8 unit ^{2}**

**Now, ****Area of region OAEO = Area of region (OABO – OEABO)**

\(\boldsymbol{\Rightarrow }\) \(8-\frac{16}{3}\) = **\(\frac{8}{3}\) unit ^{2}**

**Therefore, the total Area of shaded region = ****2****×\(\frac{8}{3}\)**** = \(\frac{16}{3}\)unit ^{2}**

** **

** **

**Q.9: ****Find the area enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** represents an **ellipse.**

**Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\\\)**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\\\)**

**\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)**

Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)

Now, the **Area enclosed by the curve ABOA:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2} y\;dx\;=\; \frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\)

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) **unit ^{2}**

Therefore, the **Area** enclosed by the **curve ABOA \(\boldsymbol{=\frac{3\pi}{2}}\)**** unit ^{2}**

**Hence, the total Area enclosed by the ellipse ABA’B’ = ****4 ****× \(\boldsymbol{\frac{3\pi}{2}}\)**** unit ^{2}**

**= 6****π unit ^{2}**

** **

** **

**Q.10: Find the area enclosed by the curve x ^{2 }+ y^{2} = 9, line x = \(2\sqrt{2}y\) and the first quadrant.**

** **

**Sol:**

** **

Equation **x ^{2 }+ y^{2 }= 9** represents a

Since, **x = \(2\sqrt{2}y\)**

Therefore (\(2\sqrt{2}y\))^{2 }+ y^{2} = 9 (for points of intersection of both the curves)

Hence, the coordinates of point **B** **= (****\(2\sqrt{2}y\)****,1)**

Now, the area of region bounded by the curve **x ^{2 }+ y^{2 }= 9**,

**i.e. Area of triangle OMB + Area under the curve MBAM.**

**Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)**

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;=\;\frac{1}{2}\times2\sqrt{2}y\times 1 =\sqrt{2}\;y\;\)unit^{2}

**Now, the Area under the curve MBAM:**

\(\boldsymbol{\Rightarrow }\) \(\int_{2\sqrt{2}}^{3}\;y\;dx\)

**Since, x ^{2} +y^{2} =9**

**Therefore, y ^{2} = 9 – x^{2}**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{9-x^{2}}}\)

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2\sqrt{2}}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\;\sin^{-1}\frac{x}{3} \right |_{2\sqrt{2}}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\;\boldsymbol{[\frac{3}{2}\times\ \sqrt{9-9}]+[\frac{9}{2}\;\sin^{-1}(1)]-[\frac{2\sqrt{2}}{2}\times \sqrt{9-8}]-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\) **unit ^{2}**

**Therefore, the Area under the curve MBAM:**

**= \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit ^{2}**

Now, total Area under the shaded region =** Area under the curve MBAM + Area of the triangle OMB**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]+\sqrt{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)** unit ^{2}**

**Hence, the required Area is given by the region OMABO:**

**\(\boldsymbol{=\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit ^{2}**

** **

** **

**Q.11: Find the area of larger part of circle x ^{2 }+ y^{2 }= 16 cut off by the line x = \(\frac{4}{\sqrt{2}}\) in the first quadrant.**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **x ^{2 }+ y^{2 }= 16 **represents a

For coordinates of point B:

\(\boldsymbol{\Rightarrow }\) \((\frac{4}{\sqrt{2}})^{2}+y^{2}=16\)

\(\boldsymbol{\Rightarrow }\) 8 + y^{2} = 16

\(\boldsymbol{\Rightarrow }\) y = \(2\sqrt{2}\)

Hence, the **coordinates** of **point** **B** are: **\(\boldsymbol{(\frac{4}{\sqrt{2}},2\sqrt{2})}\)**

**The required Area is given by the curve OMBCO:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2\sqrt{2}}y\;dx\;=\;\int_{0}^{2\sqrt{2}}\sqrt{16-x^{2}}\;dx}\)

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1} \frac{x}{4} \right |_{0}^{2\sqrt{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8\sin^{-1}\frac{1}{\sqrt{2}}]-0}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2}\times 2\sqrt{2}+8\times \frac{\pi }{4}=[4+2\pi] }\) **unit ^{2}**

**Therefore, the Area of shaded region OMBCO = [4 + 2****π****]unit ^{2}**

Let us assume two curves represented by the equation **y = f(x)** and **y = g(x) **in **[a, b]** as shown in the above figure. In this case the height of an elementary strip will be **[f(x) – g(x)] **and its width will be **dx**.

Now, Area of the elementary strip **(dA) = [f(x) – g(x)] dx**

Hence, the **total Area **of shaded region** (A) = \(\int_{a}^{b} [f(x)-g(x)]\;dx\)**

**Example – 1: Find the area enclosed between two curves whose equations are: x ^{2 }= 4y and y^{2 }= 4x.**

** **

**Sol:Based on formula given in Application of Integrals**

The Equation **x ^{2 }= 4y **represents a parabola symmetrical about y-axis and the equation

**On solving both the equations:**

\(\boldsymbol{\Rightarrow }\) \(\left ( \frac{y^{2}}{4} \right )^{2}=4y\)

\(\boldsymbol{\Rightarrow }\) y^{3} = 64 **i.e. y = 4**

Which gives x = 4. Hence, the **coordinates** of **point N** are **(4, 4)**.

Now, The Area of region enclosed by the curve** NAOBN** = Area of region enclosed by the curve **OANMO** – Area of region enclosed by the curve **OBNMO**.

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}2\sqrt{x}\;dx-\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\times \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{3}}{12} \right ]_{0}^{4}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4}{3}\times (4)^{\frac{3}{2}}-0 \right ]-\left [ \frac{64}{12}-0 \right ]= \frac{16}{3}}\) **unit ^{2}**

**Therefore, the Area of shaded region = \(\frac{16}{3}\) unit ^{2}**

**EXAMPLE – 2: Find the area enclosed by the sides of a triangle whose vertices have coordinates (1, 0) (3, 5) and (5, 4).**

** **

**Sol:Based on formula given in Application of Integrals**

**Form the above figure:**

Let, **A (1, 0), B (3, 5) and C (5, 4)** be the vertices of **triangle ABC**.

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{5\;-\;0}{3\;-\;1}\right]\)

\(\boldsymbol{\Rightarrow }\) **2y = 5x – 5**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{5x\;-\;5}{2}}\)

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-5)=(x-3)\times \left[\frac{4-5}{5-3}\right]\)

\(\boldsymbol{\Rightarrow }\) **2y – 10 = 3 – x**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{13\;-\;x}{2}}\)

**The Equation of line AC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{4-0}{5-1}\right]\)

\(\boldsymbol{\Rightarrow }\) **4y = 4x – 4**

\(\boldsymbol{\Rightarrow }\) ** y = x – 1**

Now, the** Area **of** triangle ABC = Area **under the curve** ABMA + Area **under the curve** MBCN – Area **under the curve** ACNA.**

**The Area under the curve ABMA [2y = 5x – 5]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}y\;dx\;=\;\int_{1}^{3}\frac{5x-5}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{4}-\frac{5x}{2} \right ]_{1}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}=5}\) **unit ^{2}**

Therefore**, Area **under the** curve ABMA = 5 unit ^{2}**

**The Area under the curve MBCN [2y = 13 – x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}\frac{13-x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{13x}{2}-\frac{x^{2}}{4} \right ]_{3}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{65}{2}-\frac{25}{4}-\frac{39}{2}+\frac{9}{4}=9}\) **unit ^{2}**

**Therefore, Area **under curve** MBCN = 9 unit ^{2}**

**The Area under the curve ACNA [y = x – 1]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{5}y\;dx\;=\;\int_{1}^{5}(x-1)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{2}-x \right ]_{1}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{2}-5-\frac{1}{2}+1=8}\)** unit ^{2}**

**Therefore, Area under the curve ACNA = 8 unit ^{2}**

Now, **Area** of **triangle** **ABC** = **Area** under curve **ABMA** + **Area** under curve **MBCN** – **Area** under curve **ACNA**.

**Therefore, the Area of triangle ABC = 5 + 9 – 8 = 6** **unit ^{2}**

** **

** **

** Q.1: Find the area lying above the x-axis enclosed between two curves whose equations are given as: x ^{2 }+ y^{2} = 6x and y^{2} = 3x.**

** **

**Sol:Based on formula given in Application of Integrals**

The Equation **y ^{2} = 3x** represents a

The Equation **x ^{2 }+ y^{2} = 6x** i.e.

Substituting the equation of parabola in equation of circle:

(x – 3)^{2 }+ 3x = 9 \(\boldsymbol{\Rightarrow}\) **x ^{2 }+ 9 – 6x + 3x = 9**

\(\boldsymbol{\Rightarrow}\) x^{2} – 3x = 0 i.e. x = 0 or x = 3 which gives **y = 0 and y = ****±**** 3**

Therefore, the **coordinates** of **point A** above the **x-axis** are **(3, 3)**

Now, the Area of region bounded by the curve **OQABO** = Area of region bounded by the curve **OQAMO **+ Area of region bounded by the curve **ABMA**

**The Area of region bounded by the curve OQAMO [y ^{2} = 3x]:**

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{0}^{3}y\;dx=\int_{0}^{3}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times\left [ \frac{2x^{\frac{3}{2}}}{3} \right ]_{0}^{3}=\sqrt{3}\times \frac{2}{3}\times 3^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times \frac{2}{3}\times 3\sqrt{3}=6}\)** unit ^{2}**

**Therefore, the Area of region bounded by the curve OQAMO = 6 unit ^{2}**

**Now, the Area of region bounded by the curve ABMA [(x – 3) ^{2 }+ y^{2} = 9)]:**

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{3}^{6}y\;dx=\int_{3}^{6}\sqrt{9-(x-3)^{2}}\;dx}\\\)

**Since,** \(\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\\\)

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [ \frac{x-3}{2}\sqrt{\left (3 \right )^{2}-\left ( x-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{x-3}{3} \right ]_{3}^{6}}\\\)

\(\boldsymbol{\Rightarrow \;\;\;\left [\frac {6-3}{2}\sqrt{{9}-\left (6-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{6-3}{3}\right]-\left [ \frac{x-3}{2}\;\sqrt{9-\left (3-3 \right )^{2}}\;+\frac{9}{2}\;\sin^{-1}\frac{3-3}{3}\;\right]}\\\)\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [0 +\frac{9}{2}\times \sin^{-1}1-0\right ]=\frac{9}{2}\times \frac{\pi }{2}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\frac{9\pi }{4}}\) **unit ^{2}**

**Therefore, the Area of region bounded by the curve ABMA \(=\frac{9\pi }{4}\) unit ^{2}**

**The Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA**

\(\boldsymbol{\Rightarrow}\) \(6+\frac{9\pi }{4}=\frac{24+9\pi }{4}\) unit^{2}

**Therefore, the Area of shaded region (OQABO)\(=\frac{24+9\pi }{4}\) unit ^{2}**

** **

** **

**Q.2: Find the area enclosed between two curves: 9x ^{2 }+ 9y^{2 }= 4 and (x –**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **9x ^{2 }+ 9y^{2 }= 4 . . . . . . . (1)**, represents a circle with

Equation (**x –** **\(\frac{2}{3}\) ) ^{2 }+ y^{2 }=**

**On solving equation (1) and equation (2), we will get:**

\(\\\boldsymbol{\Rightarrow }\) \((x-\frac{2}{3})^{2}+\frac{4-9x^{2}}{9}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(x^{2}+\frac{4}{9}-\frac{4x}{3}+\frac{4}{9}-x^{2}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{4}{9}=\frac{4}{3}\;x\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{x=\frac{1}{3}}\) which gives \(\boldsymbol{y=\pm \frac{1}{\sqrt{3}}}\)

Therefore, the **coordinates** of **points M** and **N** are: \((x=\frac{1}{3})\;\;(y=\pm \frac{1}{\sqrt{3}})\)

**Now, the Area enclosed by region BMONB = 2 ****×**** [Area enclosed by the curve BMOB].**

And the Area enclosed by the curve **BMOB** = Area of region enclosed by the curve **(MOPM+MPBM).**

**Now, the Area of region enclosed by the curve MPBM [****9x ^{2 }+ 9y^{2 }= 4**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{1}{3}}^{\frac{2}{3}}y\;dx=\int_{\frac{1}{3}}^{\frac{2}{3}}\sqrt{\frac{4}{9}-x^{2}}\;dx}\)

**Since,** \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-x^{2}}+\frac{4}{2\times 9}\sin^{-1}\frac{3x}{2} \right ]_{\frac{1}{3}}^{\frac{2}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac{1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\sin^{-1}(1) \right ] -\left [ \frac{1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\sin^{-1}\frac{1}{2}\right ] }\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{2}{9}\times \frac{\pi }{2} \right ]-\left [ \frac{1}{6}\times \frac{1}{\sqrt{3}}+\frac{2}{9}\times \frac{\pi }{6} \right ]=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]}\) **unit ^{2}**

**Therefore, area of region enclosed by the curve MPBM \(=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\) unit ^{2}**

**Now, the Area of region enclosed by the curve MOPM \(\left [ \left ( x-\frac{2}{3} \right )^{2}+y^{2}=\frac{4}{9} \right ]\) :**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{3}}y\;dx=\int_{0}^{\frac{1}{3}}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}\;dx}\\\)

**Since**, \(\\\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x-\frac{2}{3}}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{x-\frac{2}{3}}{\frac{2}{3}} \right ]_{0}^{\frac{1}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{\frac{1}{3}-\frac{2}{3}}{2}\sqrt{\frac{4}{9}-\left ( \frac{1}{3}-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{\frac{1}{3}-\frac{2}{3}}{\frac{2}{3}} \right]- \left [ \frac{0-\frac{2}{3}}{2}\;\sqrt{\frac{4}{9}-\left ( 0-\frac{2}{3} \right )^{2}}\;+\frac{2}{9}\;\sin^{-1}\frac{0-\frac{2}{3}}{\frac{2}{3}}\; \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\times \sin^{-1}\frac{-1}{2} \right ]- \left [ \frac{-1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\times \sin^{-1}(-1) \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6\sqrt{3}}+\frac{2}{9}\times \frac{-\pi }{6} \right ]-\left [ \frac{2}{9}\times \frac{-\pi }{2} \right ]= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]}\) **unit ^{2}**

**Therefore, the Area of region enclosed by the curve MOPM \(= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]\) unit ^{2}**

Now, the Area enclosed by the **curve BMOB** = Area of region enclosed by the curve **(MOPM+MPBM)**

\(\boldsymbol{\Rightarrow }\) \(\left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ] +\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]}\) **unit ^{2}**

Now, the Area enclosed by the curve **BMONB** = 2 [Area enclosed by the curve **BMOB**] \(=2\times \left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]\) unit^{2}

**Therefore, the Area of shaded region = \(\left [ \frac{8\pi }{27}-\frac{2}{3\sqrt{3}} \right ]\) unit ^{2}**

** **

** **

**Q.3: Find the area lying above the x-axis enclosed between two curves whose equations are given as: 4x ^{2 }+ 4y^{2} = 9 and x^{2} = 4y.**

** **

**Sol:Based on formula given in Application of Integrals**

The Equation **x ^{2} = 4y** represents a

The Equation **4x ^{2 }+ 4y^{2} = 9** i.e.

Now, on substituting the equation of parabola in the equation of circle we will get:

4(4y) + 4y^{2} =9 i.e.** 4y ^{2 }+ 16y – 9 = 0**

From the above quadratic equation: **a = 4, b = 16 and c = -9**

Substituting the values of a, b and c in **quadratic** **formula**:

\(\\\boldsymbol{\Rightarrow }\) \(\\y=\frac{-16+\sqrt{(16)^{2}-4(4\times-9)}}{2\times 4}\;and\;y=\frac{-16-\sqrt{(16)^{2}-4(4\times -9)}}{2\times 4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{256+144}}{8}\;and\;y=\frac{-16-\sqrt{256+144}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{400}}{8}\;and\;y= \frac{-16-\sqrt{400}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+20}{8}\;and\;y=\frac{-16-20}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(y = \frac{1}{2}\;\;and\;\;y = \frac{-9}{2}\)

Which gives x = \(\pm \;\sqrt{2}\) [Neglecting y = \(\frac{-9}{2}\) as it gives absurd results]

Therefore, the **coordinates** of **point** **B** are \(\left (\sqrt{2},\;\frac{1}{2} \right )\)

Now, Area of region bounded by the curve **ODCBO** = **2 ****×** (Area of region bounded by the curve **OBCO**)

Now, Area of region bounded by curve **OBCO** = (Area of region bounded by the curve **OCBMO** + Area of region bounded by the curve **OBMO**)

**Area of region bounded by the curve OCBMO [4x ^{2 }+ 4y^{2} = 9]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}\;\;dx}\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\frac{9}{4}- x^{2}}+\frac{9}{2\times 4}\sin^{-1}\frac{2\times x}{3} \right ]_{0}^{\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac {\sqrt{2}}{4}\times \sqrt{{\frac{9}{4}-2}}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right ]\\}\) **unit ^{2}**

**Therefore, Area of region bounded by the curve ABMA = \(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]\) unit ^{2}**

**Area of region bounded by the curve OBMO [x ^{2} = 4y]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{12} \right ]_{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{12}-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3\sqrt{2}}}\) **unit ^{2}**

**Therefore, the area of region bounded by the curve OBMO = \(\frac{1}{3\sqrt{2}}\) unit ^{2}**

**Now, the Area of region bounded by curve OBCO = [Area of region bounded by the curve OCBMO – Area of region bounded by the curve OBMO]**

\(\\\boldsymbol{\Rightarrow }\) **\(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-\frac{1}{3\sqrt{2}}\)**

\(\\\boldsymbol{\Rightarrow }\) **\(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

**Therefore,** **Area of region bounded by the curve** **OBCO**** = ****\(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

**Now, the Area of region bounded by the curve ODCBO = 2 ****×**** [Area of region bounded by the curve OBCO]**

\(\\\boldsymbol{\Rightarrow }\) \(2\times \left [\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\)

**Therefore, the Area of shaded region = \(\left [ \frac{1}{2\sqrt{2}}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

** **

** **

**Q.4: Find the area enclosed by the sides of a triangle whose vertices have coordinates (-2, 0) (3, 4) and (5, 2).**

** **

**Sol:Based on formula given in Application of Integrals**

**Form the above figure:**

Let, **A (1, 0), B (3, 5) and C (5, 4)** be the vertices of **triangle ABC**

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)\;=\;(x+2)\times \left[\frac{4\;-\;0}{3\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **5y = 4x + 8**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{4x\;+\;8}{5}}\)

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-4)=(x-3)\times \left[\frac{2\;-\;4}{5\;-\;3}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **2y – 8 = 6 – 2x**

\(\\\boldsymbol{\Rightarrow }\) **y = 7 – x**

**The Equation of line AC:**

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)=(x+2)\times \left[\frac{2\;-\;0}{5\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **7y = 2x + 4**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{ y= \frac{2\;x+4}{7}}\)

**Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.**

**The Area under the curve ABMA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{3}y\;dx\;=\;\int_{-2}^{3}\frac{4x\;+\;8}{5}\;dx}\\\) = \(\boldsymbol{\left [ \frac{4x^{2}}{10}+\frac{8x}{5} \right ]_{-2}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{36}{10}+\frac{24}{5}-\frac{16}{10}-\frac{-16}{5}=10}\) **unit ^{2}**

**Therefore, the Area under the curve ABMA = 10 unit ^{2}**

**The Area under the curve MBCN:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}=(7-x)\;dx}\\\) = \(\boldsymbol{\left [ 7x-\frac{x^{2}}{2} \right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{35-\frac{25}{2}- 21+\frac{9}{2}=6}\) **unit ^{2}**

**Therefore, the Area under the curve MBCN = 6 unit ^{2}**

**The Area under the curve ACNA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{5}y\;dx\;=\;\int_{-2}^{5}\frac{2x\;+\;4}{7}\;dx}\\\) = \(\boldsymbol{\left [ \frac{2x^{2}}{14}+\frac{4x}{7}\right ]_{-2}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{7}+\frac{20}{7}-\frac{4}{7}-\frac{-8}{7}=7}\) **unit ^{2}**

**Therefore, the Area under the curve ACNA = 7 unit ^{2}**

Now, **Area** of triangle **ABC** = Area under curve **ABMA** + Area under curve **MBCN** – Area under curve **ACNA**.

**Therefore, the Area of triangle ABC = 10 + 6 – 7 = 9**** unit ^{2}**

** **

** **

**Q.5: Find the area enclosed by the sides of a triangle whose equations are: y = 4x + 2, y = 3x + 2 and x = 5.**

** **

**Sol:Based on formula given in Application of Integrals**

** **

**From the above figure,**

The Equation of **side AC: y = 4x + 2 . . . . . . (1)**

The Equation of **side BC: y =3x + 2 . . . . . . (2)**

**And,** **x = 5 . . . . . . . . . . (3)**

From **equation (1) and equation (3):**

y = 4(5) + 2 = 22 **[Since, x = 5]**

Therefore, the **coordinates** of **point** **A** are **(5, 22).**

**From equation (2) and equation (3):**

y = 3(5) + 2 = 17 **[Since, x = 5]**

Therefore, the **coordinates** of **point** **B** are **(5, 17).**

**Now, substituting equation (1) in equation (2):**

3x + 2 = 4x + 2 **i.e. x = 0 and y = 2**

Therefore, the **coordinates** of **point** **C** are **(0, 2).**

Now, the **Area** of triangle **ABC** = Area enclosed by the curve **ACOMA** – Area enclosed by the curve **COMBC**.

**The Area under the curve ACOMA [****y = 4x + 2****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(4x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{50+10-0=60}\) **unit ^{2}**

**Therefore, the Area under the curve ABMA = 60 unit ^{2}.**

**The Area under the curve MBCN [y = 3x + 2]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(3x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{75}{2}+10-0=47.5}\) **unit ^{2}**

**Therefore, the Area under the curve MBCN = 47.5 unit ^{2}.**

**Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC**

\(\\\boldsymbol{\Rightarrow }\) 60 – 47.5 **= 12.5 unit ^{2}**

**Therefore, the Area of triangle ABC = 12.5 unit ^{2}**

^{ }

^{ }

**Q.6: Find the area enclosed by the curves y = x ^{2 }+ 3, y = 2x, x = 2 and x =0.**

** **

**Sol:Based on formula given in Application of Integrals**

The Equation **y = x ^{2 }+ 3** represents a

On substituting equation of line x = 2 in the equation of parabola we will get the **coordinates** of **point** **C** i.e. **(2, 7).**

On substituting **x = 2** in the equation of line **y = 2x** we will get the **coordinates** of **point B** i.e. **(2, 4).**

From the above figure,

The **Area** of region enclosed by the curve **ODCBO** = **Area** of region enclosed by the curve **ODCAO** – **Area** of region enclosed by the curve **OBAO**

**Now, the Area of region enclosed by the curve ODCAO [****y = x ^{2 }+ 3**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(x^{2}+3)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}+3x \right ]_{0}^{2}=\frac{8}{3}+6=\frac{26}{3}}\)**unit ^{2}**

**Therefore, the Area of region enclosed by the curve ODCAO = \(\frac{26}{3}\) unit ^{2}**

**Now, the Area of region enclosed by the curve OBAO [y = 2x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [x^{2} \right ]_{0}^{2}=4}\) **unit ^{2}**

**Therefore, the Area of region enclosed by the curve OBAO = 4 unit ^{2}**

Now, the **Area** of region enclosed by the **curve ODCBO** = Area of region enclosed by the **curve ODCAO** – Area of region enclosed by the **curve OBAO**

\(\Rightarrow \frac{26}{3}-4=\frac{2}{3}\) unit^{2}

**Therefore, The Area of shaded region (ODCBO) = \(\frac{2}{3}\)unit ^{2}**

^{ }

^{ }

**Q.7: Find the area enclosed between the curve y ^{2 }= 6x and line y = 3x.**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **y ^{2} = 6x** represents a parabola, symmetrical about

Now, substituting the Equation of line **y = 3x** in the equation of parabola:

(3x)^{2} = 6x **\(\Rightarrow x = \frac{2}{3}\) **which gives **y = 2**

Hence the **coordinates** of point **A** are **\((\frac{2}{3},2)\)**

**The area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve OMABO [y ^{2 }= 6x]:**

**Since, y ^{2} = 6x**

**Therefore, y = \(\sqrt{6x}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx = \int_{0}^{\frac{2}{3}}\sqrt{6x}\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{6}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{2}{3}}=\sqrt{6}\times \frac{2}{3}\times \left ( \frac{2}{3} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2} \times \sqrt{3}\times \frac{2}{3}\times 2\sqrt{2}\times \frac{1}{3\sqrt{3}}=\frac{8}{9}}\\\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO = \(\frac{8}{9}\) unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = 3x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx\Rightarrow \int_{0}^{\frac{2}{3}}3x\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\left [\frac{x^{2}}{2}\right ]_{0}^{\frac{2}{3}}=\frac{3}{2}\times \left ( \frac{2}{3} \right )^2}\) = \(\boldsymbol{\frac{2}{3}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO = \(\frac{2}{3}\) ****unit ^{2}**

**Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8}{9}-\frac{2}{3} = \frac{2}{9}\) unit^{2}

**Therefore, the Area enclosed by the curve OABO = \(\frac{2}{9}\) unit ^{2}**

**Q.1: Find the area enclosed by the curve whose equations are: y = 2x ^{2}, x = 2, x = 3 and x-axis.**

**Sol:**

The equation **y = 2x ^{2}** represents a

**Now, the Area of region enclosed by the curve ABCDA [****y = 2x ^{2}**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{3}y\;dx=\int_{2}^{3}2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2x^{3}}{3} \right ]_{2}^{3}=\left [ 18-\frac{16}{3} \right ]=\frac{38}{3}}\) **unit ^{2}**

**Therefore, the Area of shaded region \(=\frac{38}{3}\) unit ^{2}**

** **

** **

**Q.2: Find the area enclosed by the curve whose equations are: y = 5x ^{4}, x = 3, x = 7 and x-axis.**

** **

**Sol:Based on formula given in Application of Integrals**

The equation **y = 2x ^{2}** represents a

**Now, the Area of region enclosed by the curve ABCDA [****y = 5x ^{4}**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx=\int_{3}^{7}5x^{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{5}}{5} \right ]_{3}^{7}=\left [ 16807-243 \right ]=16564}\) **unit ^{2}**

**Therefore, the Area of shaded region = 16807 unit ^{2}**

** **

** **

**Q.3****: Find the area enclosed between the curve y ^{2 }= 3x and line y = 6x.**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **y ^{2} = 3x** represents a parabola, symmetrical about

Now, substituting the Equation of line **y = 6x** in the equation of parabola:

**(6x) ^{2} = 3x**

Hence the **coordinates** of point **A** are **\((\frac{1}{12},\frac{1}{2})\)**

**The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve, OMABO [y ^{2 }= 3x]:**

**Since,** **y ^{2} = 3x**

Therefore,** y = \(\sqrt{3x}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{12}} y\;dx = \int_{0}^{\frac{1}{12}}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{1}{12}}=\sqrt{3}\times \frac{2}{3}\times \left ( \frac{1}{12} \right )^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3} \times \frac{2}{3}\times \frac{1}{24\sqrt{3}}=\frac{1}{36}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO = \(\frac{1}{36}\)****unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = 6x]:**

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{48}}\)**unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO = \(\frac{1}{48}\)****unit ^{2}**

**Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{36}-\frac{1}{48} = \frac{1}{144}\)**unit ^{2}**

**Therefore, the Area enclosed by the curve OABO = \(\frac{1}{144}\)****unit ^{2}**

** **

** **

**Q.4 Find the area enclosed by the curve y = 2x ^{2} and the lines y = 1, y = 3 and the y-axis.**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **y = 2x ^{2}** represents a

** **The **Area** of the region bounded by the curve **y = 2x ^{2}**,

Now, the **Area of region AA’B’BA = 2 (Area of region ABNMA)**

**Since,** **2x ^{2} = y**

Therefore, **x = \(\sqrt{\frac{y}{2}}\)**

Thus, **the Area of region bounded by the curve ABNMA [****y = 2x ^{2}**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}x\;dy=\int_{1}^{3}\sqrt{\frac{y}{2}}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{\sqrt{2}}\times {\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{3}=\frac{\sqrt{2}}{3}\times [(3^{\frac{3}{2}})-(1)^{\frac{3}{2}}}}]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{\sqrt{2}}{3}\times (3\sqrt{3}-1) = \frac{\sqrt{2}}{3}(3\sqrt{3}-1)}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABNMA = \(\frac{\sqrt{2}}{3}(3\sqrt{3}-1)\)****unit ^{2}**

**Hence, the ****Area of region bounded by the curve AA’B’BA ****=**** 2(Area of region bounded by the curve ABNMA)****\(= \frac{2\sqrt{2}}{3}(3\sqrt{3}-1)\)****unit ^{2}**

**Q.5****: Find the area enclosed between the curve y ^{2 }= 9ax and line y = mx.**

** **

**Sol:Based on formula given in Application of Integrals**

Equation **y ^{2} = 9ax** represents a parabola, symmetrical about

Now, substituting the Equation of line **y = mx** in the equation of parabola:

9ax = (mx)^{2} i.e **x = \(\frac{9a}{m^{2}}\)** which gives **y = \(\frac{9a}{m}\)**

Hence the **co-ordinates** of point **A** are **\(\left ( \frac{9a}{m^{2}},\frac{9a}{m}\right )\)**

**The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve OMABO [y ^{2 }= 9ax]:**

**Since,** **y ^{2} = 9ax**

Therefore,** y = \(3\sqrt{ax}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{9a}{m^{2}}} y\;dx = \int_{0}^{\frac{9a}{m^{2}}}3\sqrt{ax}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\sqrt{a}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{9\;a}{m^{2}}}=3\sqrt{a}\times \frac{2}{3}\times \left ( \frac{9\;a}{m^{2}} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2\sqrt{a} \times 27\times a\sqrt{a}\times \frac{1}{m^{3}}=\frac{54\;a^{2}}{m^{3}}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO \(=\frac{54\;a^{2}}{m^{3}}\) unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = mx]:**

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{9\;a}{m^{2}}\times \frac{9\;a}{m}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{81\;a^{2}}{2\;m^{3}}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO =\(\frac{81\;a^{2}}{2\;m^{3}}\) unit ^{2}**

**Now, the Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

i.e. \(\frac{54\;a^{2}}{m^{3}}-\frac{81\;a^{2}}{2\;m^{3}} = \frac{27\;a^{2}}{2\;m^{3}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OABO = \(\frac{27\;a^{2}}{2\;m^{3}}\)unit ^{2}**

** **

**Q.6: Find the area bounded by the curve whose equation is 3x ^{2} = 4y and the line 2y – 12 = 3x.**

** **

**Sol:Based on formula given in Application of Integrals**

**Equation 3x ^{2} = 4y **represents a

The **Area** of the region bounded by parabola **3****x ^{2} = 4y** and the line

Since, the **parabola 3x ^{2} = 4y** and the

**Since,** \(\;x=\frac{2y-12}{3}\)

\(\\\boldsymbol{\Rightarrow }\) \(3\left ( \frac{2y-12}{3} \right )^{2}=4y\;\;\;i.e. \;\;\;(2y-12)^{2}=12y\)

\(\\\boldsymbol{\Rightarrow }\) 4y^{2} +144 – 48y – 12 y = 0

\(\\\boldsymbol{\Rightarrow }\) y^{2} – 15y + 36 = 0

By **splitting the middle term Method** solutions of this quadratic equation are:

y^{2} – (12+3)y + 36 = 0 \(\Rightarrow\) y(y – 12) –3(y – 12) = 0

\(\Rightarrow\) (y – 3) (y – 12) = 0

Therefore, y = 12 and y = 3 which gives x = 4 and x = -2 respectively.

Hence, the **co-ordinates** of **point A** and **point C** are** (-2, 3)** and** (4, 12) **respectively**.**

**Since,** 3x^{2} = 4y

Therefore, **y = \(\frac{3x^{2}}{4}\)**

**The Area of region bounded by the curve ABCOA = [Area of region aACba] – [Area of region OCbO + Area of region OAaO ]**

**The Area enclosed by the curve aACBa [2y – 12 = 3x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{4} y\;dx \Rightarrow \int_{-2}^{4}\frac{3x+12}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left | \frac{3x^{2}}{2}+ 12x \right |_{-2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left [ 24+48-6-(-24) \right ]=45}\) **unit ^{2}**

**The Area enclosed by the curve OAaO [****3x ^{2} = 4y**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{-2}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(0-\frac{-8}{3}) \right |=2}\) **unit ^{2}**

**The Area enclosed by the curve OCbO [****3x ^{2} = 4y**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}\;y\;dx\;\Rightarrow \int_{0}^{4} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{0}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(\frac{64}{3}-0) \right |=16}\) **unit ^{2}**

**Now, the Area of region bounded by the curve ABCOA = [ Area of region aACba ] – [ Area of region OCbO + Area of region OAaO ]**

\(\\\boldsymbol{\Rightarrow }\) **45 – [2 + 16] = 27 unit ^{2}**

**Therefore, the Area of shaded region ABCOA = 27 unit ^{2}**

** **

** **

**Q.7: Find the area enclosed by the curves {(x , y) : 6y ****≥ ****x ^{2 }and y = |x|}**

** **

**Sol:Based on formula given in Application of Integrals**

**Equation x ^{2} = 6y **represents a

The Area of the region bounded by the curve **x ^{2} = 6y** and

Now, **Area of region OAEO = OABO – OEABO**

**Since,** **x ^{2} = 6y**

Therefore, \(y=\frac{x^{2}}{6}\)

Now, **the Area of region bounded by the curve OEABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}\frac{x^{2}}{6}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{3}}{3} \right |_{0}^{6}=12}\) unit^{2}

**Therefore, the Area of region bounded by the curve OEABO = 12 unit ^{2}**

**Now, the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{6}=18}\) **unit ^{2}**

**Therefore, the Area of the region bounded by the curve OABO = 18 unit ^{2}**

**Now, ****Area of region OAEO = Area of region (OABO – OEABO)**

\(\\\boldsymbol{\Rightarrow }\) 18 – 12 = **6 unit ^{2}**

**Therefore, the total Area of shaded region = ****2 ****× 6 ****= 12 unit ^{2}**

** **

** **

**Q.8****: Find the area enclosed by the sides of a triangle whose vertices have coordinates (3, 0) (5, 8) and (7, 5).**

** **

**Sol:Based on formula given in Application of Integrals**

**Form the above figure:**

**Let,** **A (3, 0), B (5, 8) and C (7, 5)** be the vertices of **triangle ABC**.

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{8-0}{5-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y = 8x – 24

\(\boldsymbol{\Rightarrow }\) ** y=4x-12**

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-8)=(x-5)\times \left(\frac{5-8}{7-5}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y – 16 = -3x + 15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{31-3x}{2}}\)

**The Equation of line AC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{5-0}{7-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 4y=5x-15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{5x-15}{4}}\)

**Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.**

**The Area under the curve ABMA [y = 4x – 12]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}(4x-12)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}-12x\right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[50-60]-[18-36]=8}\)**unit ^{2}**

**Therefore, Area under the curve ABMA = 8 unit ^{2}**

**The Area under the curve MBCN [2y + 3x = 31]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{7}y\;dx\;=\;\int_{5}^{7}\frac{31-3x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times \left [ 31x-\frac{3x^{2}}{2}\right ]_{5}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{217}{2}-\frac{147}{4}]-[\frac{155}{2}-\frac{75}{4}]=13}\) **unit ^{2}**

**Therefore, Area under the curve MBCN = 13 unit ^{2}**

**The Area under the curve ACNA [4y = 5x – 15]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx\;=\;\int_{3}^{7}\left ( \frac{5x-15}{4} \right )dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{8}-\frac{15x}{4} \right ]_{3}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{245}{8}-\frac{105}{4} \right ]-\left [ \frac{45}{8}-\frac{45}{4} \right ]=10}\)** unit ^{2}**

**Therefore, Area under the curve ACNA = 10 unit ^{2}**

**Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA**.

**Therefore, the Area of triangle ABC = 8 + 13 – 10 = 11** **unit ^{2}**

** **

** **

**Q.9: Find the area enclosed by the sides of a triangle whose equations are: 2x – 4 = y, – 2y = -3x + 6 and x – 3y = -5.**

** **

**Sol:Based on formula given in Application of Integrals**

** **

**From the above figure:**

The Equation of **line AB: 3y = x + 5 . . . . . . (1)**

The Equation of **line BC: y = 4 – 2x . . . . . . (2)**

The Equation of **line AC: 2y = 3x – 6 . . . . . . (3)**

From **equation (1) **and** equation (2):**

3(4 – 2x) = x + 5 i.e. **x = 1**, which gives **y = 2**

Therefore, the **coordinates** of **point** **B** are **(1, 2)**

From **equation (2)** and **equation (3):**

2(4 – 2x) = 3x – 6 i.e. **x = 2** which gives **y = 0**

Therefore, the **coordinates** of **point** **C** are **(2, 0).**

From **equation** **(1)** and **equation (3):**

2y = 3(3y – 5) – 6 i.e. **y = 3** which gives **x = 4**

Therefore, the **coordinates** of **point** **A** are **(4, 3).**

Now, the **Area** of triangle **ABC** = **Area** enclosed by the curve **ABMNA** **–** **Area** enclosed by the curve **BMCB – Area **enclosed by the curve** ACNA**

**The Area under the curve ABMNA [****3y = x + 5****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{4}y\;dx\;=\;\int_{1}^{4}\frac{x+5}{3}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{6}+\frac{5x}{3} \right ]_{1}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{16}{6}+\frac{20}{3}-\frac{1}{6}-\frac{5}{3}=\frac{15}{2}}\) **unit ^{2}**

**Therefore, the Area under the curve ABMNA \(=\frac{15}{2}\) unit ^{2}**

**The Area under the curve BMCB [****y = 4 – 2x****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{2}y\;dx\;=\;\int_{1}^{2}(4-2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 4x-x^{2} \right ]_{1}^{2}}\)

\(\\\boldsymbol{\Rightarrow }\) [8 – 4] – [4 – 1] **=1 unit ^{2}**

**Therefore, the Area under the curve MBCN = 1 unit ^{2}**

**The Area under the curve ACNA [****2y = 3x – 6****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{4}y\;dx\;=\;\int_{2}^{4}\frac{3x-6}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{4}-\frac{6x}{2} \right ]_{2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3\times 16}{4}-\frac{6\times 4}{2} \right ]- \left [ \frac{3\times 4}{4}-\frac{6\times 2}{2} \right ]=3}\)** unit ^{2}**

**Therefore, the Area under the curve ACNA = 3 unit ^{2}**

The **Area** of triangle **ABC** = **Area** enclosed by the curve **ABMNA** **–** **Area** enclosed by the curve **BMCB – Area **enclosed by the curve** ACNA**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{15}{2}-1-3 = 3.5\) unit ^{2}**

**Therefore, the Area of triangle ABC = 3.5 unit ^{2}**

** **

** **

**Q.10: Find the area enclosed by the curve 2x ^{2 }= y and the line y = 2x + 12 and x – axis.**

** **

**Sol:Based on formula given in Application of Integrals**

**Equation 2x ^{2} = y **represents a

The **Area** of the region bounded by parabola **2****x ^{2} = y** and the line

Since, the **parabola 2x ^{2} = y** and the

**Since, y = 2x + 12**

\(\\\boldsymbol{\Rightarrow }\) 2x^{2 }= (2x+12)

\(\\\boldsymbol{\Rightarrow }\) **x ^{2} – x – 6 = 0**

By **splitting the middle term Method** solutions of this quadratic equation are:

x^{2} – (3 – 2)x – 6 = 0 \(\Rightarrow\) x(x – 3) +2(x – 3) = 0

\(\Rightarrow\) (x – 3) (x + 2) = 0

Therefore, **x = 3** and **x = -2** which gives **y = 18** and **y = 8** respectively.

Hence, the **co-ordinates** of **point E** and **point A** are** (3, 18)** and** (-2, 8) **respectively**.**

**Since,** **2x ^{2} = y**

**Therefore,** **y = 2x ^{2}**

**The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]**

**The Area enclosed by the curve ACOA [ ****2x ^{2 }= y **

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} 2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{2\;x^{3}}{3}\right |_{-2}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left |(0-\frac{-16}{3}) \right |=\frac{16}{3}}\) **unit ^{2}**

**The Area enclosed by the curve ABC [ ****y = 2x + 12 ****] :**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area\;of\;\Delta ABC=\frac{1}{2}\times Base\times Altitude}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times |BC|\times |AC|=\frac{1}{2}\times \left | 4 \right |\times |8|=16}\) unit^{2}

**The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]**

\(\Rightarrow \frac{16}{3}+16=\frac{64}{3}\) unit^{2}

**Therefore, the Area of shaded region ABCOA \(=\frac{64}{3}\)unit ^{2}**

** **

** **

**Q.11: Plot the curve y = |x + 4| and hence evaluate \(\int_{-9}^{0}|x+4|\;dx\)**

** **

**Sol:Based on formula given in Application of Integrals**

From the given equation the corresponding values of x and y are given in the following table.

X |
-9 | -8 | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

Y |
5 | 4 | 3 | 2 | 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Now, on using these values of x and y, we will plot the graph of **y = |x + 4|**

** **

From the above graph,** the required Area = the Area enclosed by the curve ABCA + the Area enclosed by the curve CDOC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-9}^{0}|x+4|\;dx+\int_{-9}^{-4}(x+4)\;dx+\int_{-4}^{0}(x+4)\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-9}^{-4}+\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-4}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | 8-16-\frac{81}{2}+36 \right |+\left | 0-(8-16) \right |=\left | \frac{-25}{2} \right |+8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{41}{2}}\)** unit ^{2}**

**Therefore, the area of shaded region = \(\frac{41}{2}\) unit ^{2}**

**Q.12: Find the area enclosed by the curve y = sin x between 0 ****≤ ****x ****≤ 2π**

**Sol:Based on formula given in Application of Integrals**

From the above figure, the required Area is represented by the curve **OABCD**.

**Now, the Area bounded by the curve OABCD = the Area bounded by the curve OABO + the Area bounded by the curve BCDB**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\pi } sin(x)\;dx+\left | \int_{\pi }^{2\pi } sin(x)\;dx\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x)\right ]_{0}^{\pi }+\left | \left [ -cos (x) \right ] _{\pi }^{2\pi }\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[-cos(\pi )+cos(0)]+\left | [-cos(2\pi )+cos(\pi )] \right |}\\\)

\(\boldsymbol{\Rightarrow }\) **[1+1] + [ |– 1 – 1| ] unit ^{2}**

**Therefore, the area of shaded region = 4 unit ^{2}**

** **

** **

**Q.13: ****Find the area of smaller region enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) and the line \(\frac{x}{2}+\frac{y}{3}=1\)**

** **

**Sol:Based on formula given in Application of Integrals**

The Equation **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** represents an **ellipse.**

The Equation **\(\frac{x}{2}+\frac{y}{3}=1\)** represents a **line** with **x and y** **intercepts** as **2 and 3** respectively.

** **

**Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\)**

**\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)**

Therefore, the **Area** of **smaller** **region** enclosed by the Ellipse **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** and the line **\(\frac{x}{2}+\frac{y}{3}=1\)** is represented by **curve ACBA**

**Now, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – Area enclosed by the curve ABOA**

**Now, the Area enclosed by the curve ACBOA:**

**Since, \(\ \int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve ACBOA = \(\boldsymbol{\frac{3\pi}{2}}\)unit ^{2}**

**Now, the Area enclosed by the curve ABOA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area \;of\;\Delta ABO = \frac{1}{2}\times AO\times BO}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times 2\times 3=3}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve ABOA = 3 unit ^{2}**

**Since, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – the Area enclosed by the curve ABOA**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)\)**unit ^{2}**

**Therefore, the Area of shaded region \(=\frac{3}{2}(\pi -2)\) unit ^{2}**

^{ }

** **

**Q.14: Find the area enclosed by the curve |x| + |y| = 2, by using the method of integration.**

** **

**Sol: Based on formula given in Application of Integrals**

Equation **|x| + |y| = 2** represent a region bounded by the lines:

**x + y = 2 . . . . . . (1)**

**x – y = 2 . . . . . . (2)**

**-x + y = 2 . . . . . . (3)**

**-x – y = 2 . . . . . . (4)**

**From equations (1), (2), (3) and (4)** we conclude that the curve intersects **x-axis** and **y-axis** axis at points **A (0, 2), B (2, 0), C (0, -2)** and **D (-2, 0) respectively.**

**From the above figure:**

Since, the curve is symmetrical to **x-axis** and **y-axis**. Therefore, the **Area** of region bounded by the curve **ABCDA** = **4** × **Area** of region bounded by the curve **ABOA**

**Now, the Area of region bounded by the curve ABOA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=4\int_{0}^{2}(2-x)dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2x-\frac{x^{2}}{2} \right ]_{0}^{2}=(4-2)=2}\\\)** unit ^{2}**

**Therefore, the Area of region bounded by the curve ABOA = 2unit ^{2}**

**Since, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA**

**Therefore, the Area of region bounded by the curve ABCDA =** (2 × 4) **= 8 unit ^{2}**

**Therefore, the Area of shaded region = 8 unit ^{2}**

** **

** **

**Q.15: Find the area which is exterior to curve x ^{2} = 2y and interior to curve x^{2 }+ y^{2} = 15.**

** **

**Sol: Based on formula given in Application of Integrals**

The Equation **x ^{2} = 2y** represents a

The Equation **x ^{2 }+ y^{2} = 15** represents a

Now, on substituting the equation of parabola in the equation of circle we will get:

(2y) + y^{2} = 15 i.e.** y ^{2 }+ 2y – 15 = 0**

Now, by **splitting the middle term method** solutions of this quadratic equation are:

y^{2} + (5 – 3)y – 15 = 0 \(\Rightarrow\) y(y + 5) – 3(y +5) = 0

\(\Rightarrow\) (y – 3) (y + 5) = 0

Neglecting y = -5 ** [gives absurd values of x]**

Therefore, **y = 3 **which gives **x =** **\(\pm \sqrt{6}\)**

Hence, the **coordinates** of **point** **B and point D** are **(\(\sqrt{6}\), 3) (\(-\sqrt{6}\)****, 3****)** respectively.

**Now, the Area of region bounded by the curve BAC’A’DOB = 2 ****× ****(Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB+ Area of region bounded by the curve OAC’O)**

**Area of region bounded by the curve BAMB [ x ^{2 }+ y^{2} = 15 ]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\sqrt{6}}^{\sqrt{15}}y\;dx=\int_{\sqrt{6}}^{\sqrt{15}}\sqrt{\left ( \sqrt{15} \right )^{2}-x^{2}}\;\;dx}\\\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{15- x^{2}}+\frac{15}{2}\sin^{-1}\frac{x}{\sqrt{15}} \right ]_{\sqrt{6}}^{\sqrt{15}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac {\sqrt{15}}{2}\times \sqrt{{15-15}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{15}}{\sqrt{15}} \right ]-}\\\) \(\\\boldsymbol{\left [\frac {\sqrt{6}}{2}\times \sqrt{{15-6}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{6}}{\sqrt{15}}\right]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{15\pi }{4} \right ]-\left [\frac{3\sqrt{6}}{2}+\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}\right]}\\\)

**\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{15\pi }{4}=\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5} \right ]}\)unit ^{2}**

**Area of region bounded by the curve OBMO [ x ^{2} = 2y ]:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{6}}y\;dx=\int_{0}^{\sqrt{6}}\frac{x^{2}}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{6} \right ]_{0}^{\sqrt{6}}=\frac{6\sqrt{6}}{6}-0}\\\) = \(\boldsymbol{\sqrt{6}}\) **unit ^{2}**

**Area of region bounded by the curve OAC’O:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( \sqrt{15} \right )^{2} }{4}}\\\) = \(\boldsymbol{\frac{15\pi }{4}}\)unit^{2}

Now, the Area of region bounded by the curve **BAC’A’DOB** = **2 ****×** (**Area** of region bounded by the curve **OBMO + Area **of region bounded by the curve** BAMB +** **Area** of region bounded by the curve **OAC’O**)

**Therefore, the Area of region bounded by the curve BAC’A’DOB:**

\(\boldsymbol{\Rightarrow }\) \(2\left [ \frac{15\pi }{4}-\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}+\sqrt{6} +\frac{15\pi }{4}\right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\\\)**unit ^{2}**

**Therefore, the Area of shaded region: \(\boldsymbol{ \left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\)unit ^{2}**

** **

** **

**Q.16: Find the area enclosed by the curve y = x ^{3}, x-axis and the lines x = -2 and x = 2.**

** **

**Sol. Based on formula given in Application of Integrals**

The Equation **y = x ^{3}** represents a

**From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}x^{3}\;dx+\left | \int_{-2}^{0}x^{3}\;dx \right |}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{4}}{4} \right ]_{0}^{2}+\left | \left [ \frac{x^{4}}{4} \right ]_{-2}^{0} \right |}\)

\(\\\boldsymbol{\Rightarrow }\) **4 – 0 + 0 + 4 =16 unit ^{2}**

**Therefore, the Area of shaded region = 16 unit ^{2}**

** **

** **

**Q.17: Find the area enclosed by the curve y = x|x|, y – axis and the lines y = -1 and y = 3.**

** **

**Sol: Based on formula given in Application of Integrals**

**Now, y = x|x| is equal to [y = x ^{2}] when x > o**

**And y = x|x| is equal to [y = -x ^{2}] when x < o**

** **

**From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}x^{2}\;dx+\left | \int_{-1}^{0}-x^{2}\;dx \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}\right ]_{0}^{3}+\left | \left [ \frac{x^{3}}{3} \right ]_{-1}^{0} \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{9+\frac{1}{3}=\frac{28}{3}}\)**unit ^{2}**

**Therefore, the area of shaded region \(=\frac{28}{3}\)unit ^{2}**

** **

** **

**Q.18: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and y – axis, when [0 ****≤ x ≤ \(\frac{\pi }{2}\)****].**

** **

**Sol: Based on formula given in Application of Integrals**

**y = Cos(x) . . . . . . . . (1)**

**y = Sin(x) . . . . . . . . . (2)**

Now, **from equation (1) and equation (2):**

**Cos (x) = Sin (x)** \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\)

\(\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\) \(\Rightarrow\) **x = \(\frac{\pi }{4}\)**

**Therefore, the coordinates of point A are:** **\(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)
**

** **

**Now, from the above figure:**

**The required Area =** **Area** enclosed by the **curve ADMA** **+** **Area** enclosed by the **curve AMOA**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\\\)

**Since, \(\int \sin^{-1}y\;dy= y\sin^{-1}y+\sqrt{1-y^{2}}\\\)**

**And, \(\\\int \cos^{-1}y\;dy= y \cos^{-1}y-\sqrt{1-y^{2}}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ y\sin^{-1}y+\sqrt{1-y^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}+\left [ y\cos^{-1}y-\sqrt{1-y^{2}} \right ]_{\frac{1}{\sqrt{2}}}^{1}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{1}{\sqrt{2}}\times \frac{\pi }{4}+\frac{1}{\sqrt{2}}-(0+1) \right ]+\left [ 0-\left ( \frac{1}{\sqrt{2}}\times \frac{\pi }{4}-\frac{1}{\sqrt{2}} \right ) \right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}-1 \right ]}\)**unit ^{2}**

**Therefore, the Area of shaded region = \(\left [\sqrt{2}-1 \right ]\)unit ^{2}**

^{ }

^{ }

**Q.20: Find the area which is exterior to curve y ^{2} = 6x and interior to curve x^{2 }+ y^{2} = 16.**

** **

**Sol: Based on formula given in Application of Integrals**

The Equation **y ^{2} = 6x** represents a

The Equation **x ^{2 }+ y^{2} = 16** represents a

Now, on substituting the equation of parabola in the equation of circle we will get:

x^{2} + (6x) = 16 i.e.** x ^{2 }+ 6x – 16 = 0**

Now, by **splitting of middle term method** solutions of this quadratic equation are:

x^{2} + (8 – 2)x – 16 = 0 \(\Rightarrow\) x(x + 8) – 2(x + 8) = 0

\(\Rightarrow\) (x – 2) (x + 8) = 0

Neglecting x = -8 [gives absurd values of y]

Therefore, **x = 2 **which gives **y = \(\pm 2\sqrt{3}\)**

Hence, the **coordinates** of **point** **B and point D** are **(****2, \(2\sqrt{3}\)****) (2, \(-2\sqrt{3}\))** respectively.

**Now, the Area of region bounded by the curve PBA’B’NOP** **=** **2 ****×** (**Area** of region bounded by the curve **BPMOB – Area **of region bounded by the curve** POMP +** **Area** of region bounded by the curve **BOA’B**)

**Area of region bounded by the curve BPMOB [x ^{2 }+ y^{2} = 16]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{\left ( 4\right )^{2}-x^{2}}\;\;dx}\\\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{16- x^{2}}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2}{2}\times \sqrt{{16-4}}+8\times \sin^{-1}\frac{1}{2} \right ]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\sqrt{3}+\frac{8\times \pi }{6} \right ]=\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]}\\\)**unit ^{2}**

**Therefore, Area of region bounded by the curve BPMOB = \(\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]\)unit ^{2}**

**Area of region bounded by the curve POMP [y ^{2} = 6x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{6x}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\left [ \frac{\sqrt{6}\times 2}{3}\times x^\frac{3}{2} \right ]_{0}^{2}=\frac{2\sqrt{6}}{3}\times 2\sqrt{2}-0\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8\sqrt{3}}{3}\)**unit ^{2}**

**Therefore, Area of region bounded by the curve POMP : \(\frac{8\sqrt{3}}{3}\)unit ^{2}**

**Area** of region bounded by the curve **BOA’B:**

\(\\\boldsymbol{\Rightarrow }\) **4****π**** unit ^{2}**

Therefore, **Area** of region bounded by the curve **BOA’B = 4****π**** unit ^{2}**

Now, the **Area** of region bounded by the curve **PBA’B’NOP** = **2 ****× **(**Area** of region bounded by the curve **BPMOB – Area **of region bounded by the curve** POMP +** **Area** of region bounded by the curve **BOA’B**)

**Therefore, **the **Area** of region bounded by the curve **PBA’B’NOP:**

\(\\\boldsymbol{\Rightarrow }\) \(2\left [2\sqrt{3}+\frac{4\pi }{3}-\frac{8\sqrt{3}}{3}+4\pi \right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{32\pi }{3}-\frac{4\sqrt{3}}{3} \right ]=\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]}\\\)**unit ^{2}**

**Therefore, the Area of shaded region = \(\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]\)unit ^{2}**

** **

** **

**Q.21: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and x – axis, when [0 ****≤ x ≤ **** ].**

** **

**Sol:Based on formula given in Application of Integrals**

**y = Cos(x) . . . . . . . . (1)**

**y = Sin(x) . . . . . . . . . (2)**

Now, from **equation (1)** and **equation (2):**

**Cos (x) = Sin (x)** \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\\\)

\(\\\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\\\) \(\\\Rightarrow\) **x = \(\frac{\pi }{4}\)**

**Therefore, the coordinates of point A are:** **\(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)**

**Now, from the above figure:**

**The required Area =** **Area** enclosed by the **curve AONA** **+** **Area** enclosed by the **curve ANBA**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x) \right ]_{0}^{\frac{\pi }{4}}+[sin (x)]_{\frac{\pi }{4}}^{\frac{\pi }{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos\frac{\pi }{4}+cos(0)+sin\frac{\pi }{2}-sin\frac{\pi }{4}\right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}} \right ]=[2-\sqrt{2}\;]}\)**unit ^{2}**

**Therefore, the Area of shaded region \( = [2-\sqrt{2}\;]\)unit ^{2}**