NCERT Exemplar for Class 12 Maths - ApplicationOf Integrals - Free PDF Download

Solved Examples

Short Answers

1. Find the area of the curve $y=\sin x$ between 0 and π.

Ans: We have,

Area OAB $={\int }_0^{\pi }ydx={\int }_0^{\pi }\sin xdx$ 

$={|-\cos x|}_0^{\pi }$

$\text{=cos0 }\text{ -- }\text{ cos }\pi \text{ = }\text{ 2 }\text{ sq }\text{ units}\text{.}$

2. Find the area of the region bounded by the curve $a{y}^2=x^3$, the y-axis and the lines y = a and y = 2a.

Ans: We have

Area BMNC $$=\underset{a}{\overset{2a}{\int }}\text{xdy}=\underset{a}{\overset{2a}{\int }}{\text{a}}^{\frac{\text{1}}{\text{3}}}{\text{y}}^{\frac{\text{2}}{\text{3}}}\text{dy}$$

$\text{=}\frac{\text{3}{\text{a}}^{\frac{\text{1}}{\text{3}}}}{\text{5}}{\left|{\text{y}}^{\frac{\text{5}}{\text{3}}}\right|}_{\text{a}}^{\text{2a}}$

$\text{=}\frac{\text{3}{\text{a}}^{\frac{\text{1}}{\text{3}}}}{\text{5}}{\left|{\left(\text{2a}\right)}^{\frac{\text{5}}{\text{3}}}\text{-}{\text{a}}^{\frac{\text{5}}{\text{3}}}\right|}_{}^{}$

$\text{=}\frac{\text{3}{\text{a}}^{\frac{\text{1}}{\text{3}}}{\text{a}}^{\frac{\text{5}}{\text{3}}}}{\text{5}}{\left|{\text{2}}^{\frac{\text{5}}{\text{3}}}\text{-1}\right|}_{}^{}$

$\text{=}\frac{\text{3}{\text{a}}^{\text{2}}}{\text{5}}\left|\text{2}\text{.}{\text{2}}^{\frac{\text{1}}{\text{3}}}\text{-1}\right|$sq. units

3. Find the area of the region bounded by the parabola $y^2=2x$ and the straight line x – y = 4.

Ans: 

The intersecting points of the given curves are obtained by solving the equations x – y = 4 and $y^2=2x$ for x and y. We have $y^2=8+2y$ i.e., (y – 4) (y + 2) = 0 which gives y = 4, –2 and x = 8, 2. Thus, the points of intersection are (8, 4), (2, –2). Hence

Area $$=\underset{-2}{\overset{4}{\int }}\left(\text{4+y-}\frac{{\text{y}}^{\text{2}}}{\text{2}}\right)\text{dy}$$

$\text{=}{\left|\text{4y+}\frac{{\text{y}}^{\text{2}}}{\text{2}}\text{-}\frac{{\text{y}}^{\text{3}}}{\text{6}}\right|}_{\text{-2}}^{\text{4}}\text{=18 }\text{ sq}\text{. }\text{ units}$

4. Find the area of the region bounded by the parabolas ${\text{y}}^{\text{2}}$ = 6x and $x^2$ = 6y.

Ans: 

The intersecting points of the given parabolas are obtained by solving these

equations for x and y, which are 0(0, 0) and (6, 6). Hence

Area OABC $={\int }_0^6\left(\sqrt{\text{6x}}\text{-}\frac{{\text{x}}^{\text{2}}}{\text{6}}\right)\text{dx=}{\left|\text{2}\sqrt{\text{6}}\frac{{\text{x}}^{\text{3/2}}}{\text{3}}\text{-}\frac{{\text{x}}^{\text{3}}}{\text{18}}\right|}_{\text{0}}^{\text{6}}$

$=|2\sqrt{6}\frac{6^{3/2}}{3}-\frac{6^3}{18}|=12sq.units$

5. Find the area enclosed by the curve x = 3 cost, y = 2 sint.

Ans: Eliminating t as follows: x = 3 cost, y = 2 sin t ⇒ cos t = $\frac{x}{3}$ , sin t = $\frac{y}{2}$ , we obtain

$\frac{x^2}{9}+\frac{y^2}{4}=1$,

Which is the equation of an ellipse.

From figure we get:

The required area $=4{\int }_0^3\frac{2}{3}\sqrt{9-x^2}dx$

$=\frac{8}{3}{[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}si{n}^{-1}\frac{x}{3}]}_0^3=6\pi sq.units$

Long Answers

6. Find the area of the region included between the parabola $y=\frac{3x^2}{4}$ and the line 3x – 2y + 12 = 0. 

Ans: Solving the equations of the given curves $y=\frac{3x^2}{4}$ and 3x – 2y + 12 = 0, we get 3$x^2$ – 6$x$ – 24 = 0 ⇒ (x – 4) (x + 2) = 0

⇒ x = 4, x = –2 which give y = 12, y = 3 

From Fig, the required area = area of ABC 

$={\int }_{-2}^4\left(\frac{12+3x}{2}\right)dx-{\int }_{-2}^4\left(\frac{3x^2}{4}\right)dx$

$={(6x+\frac{3x^2}{4})}_{-2}^4-{\left|\frac{3x^3}{12}\right|}_{-2}^4=27sq.units$

7. Find the area of the region bounded by the curves x = a$t^2$ and y = 2at between the ordinates corresponding to t = 1 and t = 2.

Ans: Given that x = a$t^2$ ...(i), y = 2at ...(ii) ⇒ t = $\frac{y}{2a}$ putting the value of t in (i), we get $y^2$ = 4ax 

Putting t = 1 and t = 2 in (i), we get x = a, and x = 4a 

Required area = 2 area of ABCD 

$=2{\int }_a^{4a}ydx=2\times 2{\int }_a^{4a}\sqrt{ax}dx$

$=8\sqrt{a}{\left|\frac{x^{3/2}}{3}\right|}_a^{4a}=\frac{56}{3}{a}^{2}sq.units$

8. Find the area of the region above the x-axis, included between the parabola $y^2$ = ax and the circle $x^2$ + $y^2$ = 2ax.

Ans: Solving the given equations of curves, we have $x^2$ + ax = 2ax or x = 0, x = a, which give y = 0, y = ± a

From fig -

Area ODAB $={\int }_0^a(\sqrt{2ax-x^2}-\sqrt{ax})dx$

Let x = 2a $si{n}^2\theta$. Then dx = 4a sinθ cosθ dθ and x = 0, ⇒ θ = 0, x = a ⇒ θ = $\frac{\pi }{4}.$

Again, ${\int }_0^a\left(\sqrt{2ax-x^2}\right)dx$

$={\int }_0^{\frac{\pi }{4}}(2a.sin\theta .cos\theta )(4a.sin\theta .cos\theta )d\theta$

$=a^2{\int }_0^{\frac{\pi }{4}}(1-cos4\theta )d\theta =a^2{(\theta -\frac{sin4\theta }{4})}_0^{\frac{\pi }{4}}=\frac{\pi }{4}{a}^2$

Furthermore,

${\int }_0^a\left(\sqrt{ax}\right)dx=\sqrt{a}\frac{2}{3}{\left(x^{\frac{3}{2}}\right)}_0^a=\frac{2}{3}{a}^2$

Thus the required area = $\frac{\pi }{4}{a}^2$ – $\frac{2}{3}{a}^2$ = $a^2\left(\frac{\pi }{4}\frac{2}{3}\right)$ sq units

9. Find the area of a minor segment of the circle $x^2+y^2=a^2$ cut off by the line x = $\frac{a}{2}$. 

Ans: Solving the equation $x^2+y^2=a^2$ and  x = $\frac{a}{2}$, we obtain their points of intersection which are $(\frac{a}{2},\sqrt{3}\frac{a}{2})and(\frac{a}{2},-\sqrt{3}\frac{a}{2}).$

Hence, from fig, we get -

Required area = 2 $\times$ area of OAB $=2{\int }_{a/2}^a\sqrt{a^2-x^2}dx$

$=2{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}]}_{a/2}^a$

$=2{[\frac{a^2}{2}\frac{\pi }{2}-\frac{a}{4}.a.\frac{\sqrt{3}}{2}-\frac{a^2}{2}.\frac{\pi }{6}]}_{}^{}$

$=\frac{a^2}{12}(6\pi -3\sqrt{3}-2\pi )$

$=\frac{a^2}{12}(4\pi -3\sqrt{3})sq.units$

Objective type questions

Choose the correct answer from the given four options in each of the Examples 10 to 12.

10. The area enclosed by the circle $x^2+y^2=2$ is equal to

(A) 4$\pi$ sq. units                         

(B) $2\sqrt{2}\pi$ sq. units

(C) 4${\pi }^2$ sq. units                      

(D) $2\pi$ sq. units   

Ans: Correct answer is (D); since Area $=4{\int }_0^{\sqrt{2}}\sqrt{2-x^2}dx$

$=4{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}]}_0^{\sqrt{2}}=2\pi sq.units$

11. The area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to

(A) ${\pi }^{2}ab$        

(B) $\pi ab$          

(C) $\pi a^{2}b$           

(D) $\pi a{b}^2$

Ans: Correct answer is (B); since Area $=4{\int }_0^a\frac{b}{a}\sqrt{a^2-x^2}dx$

$=4\frac{b}{a}{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}]}_0^a=\pi absq.units$

12. The area of the region bounded by the curve $y=x^2$ and the line y = 16.

(A) $\frac{32}{3}$          

(B) $\frac{256}{3}$             

(C) $\frac{64}{3}$          

(D ) $\frac{128}{3}$

Ans: Correct answer is (B); since Area $=2{\int }_0^{16}\sqrt{y}dy$

$=2{\left(\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right)}_0^{16}=\frac{4}{3}({16}^{\frac{3}{2}}-0)$

$=\frac{4}{3}(4^{2\times \frac{3}{2}}-0)$

$=\frac{4^4}{3}=\frac{256}{3}sq.units$

Fill in the blanks in each of the Examples 13 and 14.

13. The area of the region bounded by the curve $x=y^2$ , y-axis and the line y=3 and y = 4 is _______.

Ans: since Area $={\int }_3^{4}xdy={\int }_3^4{y}^{2}dy$

$={\left(\frac{y^3}{3}\right)}_3^4={(\frac{4^3}{3}-\frac{3^3}{3})}_{}^{}$

$=(\frac{64}{3}-\frac{27}{3})=\frac{37}{3}sq.units$

14. The area of the region bounded by the curve y =$x^2+x$, x-axis and the line x = 2 and x = 5 is equal to ________. 

Ans: since Area $={\int }_2^{5}ydx={\int }_2^5(x^2+x)dx$

$={\left(\frac{x^3}{3}\right)}_2^5+{\left(\frac{x^2}{2}\right)}_2^5={(\frac{5^3}{3}-\frac{2^3}{3})}_{}^{}+(\frac{5^2}{2}-\frac{2^2}{2})$

$=(\frac{125}{3}-\frac{8}{3})+(\frac{25}{2}-\frac{4}{2})$

$=\left(\frac{117}{3}\right)+\left(\frac{21}{2}\right)=\frac{234+63}{6}=\frac{297}{6}sq.units$

Short Answer Questions

1. Find the area of the region bounded by the curves $y^2$ = 9x, y = 3x.

Ans: We have $y^2$ = 9x, y = 3x.

         $\Rightarrow {\left(3x\right)}^2=9x$

         $\Rightarrow 9x^2-9x=0$

         $\Rightarrow 9x(x-1)=0$

         $\Rightarrow$ x = 0,1

         Therefore, required area = ${\int }_0^1\sqrt{9x}dx-{\int }_0^{1}3xdx$

         $\Rightarrow$ 3 ${\int }_0^1{x}^{\frac{1}{2}}$dx - 3${\int }_0^{1}xdx$

         $\Rightarrow$ 3 ${\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1$-  3 ${\left[\frac{x^2}{2}\right]}_0^1$

         $\Rightarrow$ 3 ($\frac{2}{3}-0)$ - 3 ($\frac{1}{2}-0)$

         $\Rightarrow$ 2- $\frac{3}{2}$ = $\frac{1}{2}$ units

2. Find the areaof the region bounded by the parabola $y^2$ = 2px,$x^2$ = 2py.

Ans: We have $y^2$ = 2px, $x^2$ = 2py.

         y = $\sqrt{2px}$

        $\Rightarrow$ $x^2$= 2p$\sqrt{2px}$

        $\Rightarrow$ $x^4$ = 4 $p^2$(2px)

        $\Rightarrow$ $x^4$ = 8$p^3$x

        $\Rightarrow$ $x^4$ - 8$p^3$x = 0

        $\Rightarrow$ $x(x^3$ - 8$p^3$) =0

        $\Rightarrow$ x =0, 2p

      Required area = ${\int }_0^{2p}\sqrt{2px}$ dx - ${\int }_0^{2p}\frac{x^2}{2p}$ dx 

      = $\sqrt{2p}$ ${\int }_0^{2p}\sqrt{x}$ dx - $\frac{1}{2p}{\int }_0^{2p}{x}^2$ dx

     =$\sqrt{2p}{\left[\frac{2{\left(x\right)}^{\frac{3}{2}}}{3}\right]}_0^{2p}$-  $\frac{1}{2p}{\left[\frac{{\left(x\right)}^3}{3}\right]}_0^{2p}$

     = $\sqrt{2p}[\frac{2{\left(2p\right)}^{\frac{3}{2}}}{3}-0]$-  $\frac{1}{2p}$[$\frac{{\left(2p\right)}^3}{3}$ -0]

    = $\sqrt{2p}$($\frac{2}{3}.2\sqrt{2}$ $p^{\frac{3}{2}}$)- $\frac{1}{2p}$($\frac{1}{3}$.8 $p^3)$ 

= $\frac{4\sqrt{2}}{3}$.$\sqrt{2}{\text{p}}^{\text{2}}$- $\frac{8}{6}{p}^2$

= $\frac{8}{6}{p}^2$

= $\frac{4}{6}{p}^2$ sq unit 

 3. Find the area of the region bounded by the curve y = $x^3$ and y = x + 6 and 

     x = 0.

Ans: We have, y = $x^3$ and y = x + 6 and x = 0.

   Therefore, $x^3$ = x + 6 

   $\Rightarrow$ $x^3$ - x = 6

  $\Rightarrow$ $x^3$ - x - 6 = 0

  $\Rightarrow$ $x^2(x-2)$+ 2x (x-2) + 3 (x-2)= 0 

  $\Rightarrow$ (x-2) ($x^2$+ 2x +3) = 0

  $\Rightarrow$ x = 2, With two imaginary points

Therefore, required area of shaded region= ${\int }_0^2(x+6-x^3$) dx

= $[\frac{x^2}{2}+6x-{\frac{x^4}{2}}_{}^{}{]}_0^2$

= [ $\frac{2^2}{2}$ + 6(2) - $\frac{2^4}{2}$-0]

= [2 + 12 -4] = 10 sq units

4. Find the area of the region bounded by the curve $y^2$ = 4x, $x^2$= 4y.

Ans: Given equations of curves are

         $y^2$ = 4x ---------- (i)

         and $x^2$= 4y.------- (ii)

       $\Rightarrow$ $\frac{x^2}{4}$ = 4x

$\Rightarrow$ $\frac{x^2}{4.4}$ = 4x

$\Rightarrow$ $x^4$= 64 x

$\Rightarrow$ $x^4$- 64 x= 0

$\Rightarrow$ x($x^3$- 64) =0 

$\Rightarrow$ x ($x^3$-$4^3$)=0

$\Rightarrow$ x = 4 ,0

 Therefore, area of shaded region A = ${\int }_0^4$($\sqrt{4x}$  - $\frac{x^2}{4})dx$

     = ${\int }_0^4$($2\sqrt{x}$  - $\frac{x^2}{4})dx$

    = ${[\frac{2x^{\frac{3}{2}}.2}{3}-\frac{1}{4}.\frac{x^3}{3}]}_0^4$

    = $\frac{32}{3}$-$\frac{16}{3}$ =  $\frac{16}{3}$ sq units

d

5. Find the area of the region included between $y^2$ = 9x and y = x

Ans: We have , $y^2$ = 9x and y = x

        $\Rightarrow$ $x^2$ = 9 x

        $\Rightarrow$ $x^2$ - 9 x =0 

        $\Rightarrow$ $x(x$ - 9 ) =0

        $\Rightarrow$ x = 0,9

Therefore, area of shaded region A = ${\int }_0^9$($\sqrt{9x}$  - x ) dx 

=  ${\int }_0^9$(3$\sqrt{x}$  - x ) dx 

= ${\int }_0^9$(3$\sqrt{x}$  ) dx - ${\int }_0^9$( x ) dx

= ${[\frac{6x^{\frac{3}{2}}}{3}-\frac{x^2}{2}]}_0^9$

= ( 6.$\frac{9^{\frac{3}{2}}}{3}$-$\frac{81}{2}-0)$

= 54 -$\frac{81}{2}$ 

= $\frac{108-81}{2}$

= $\frac{27}{2}$ sq units

6. Find the area of the region enclosed by the parabola $x^2$ = y and the line 

y = x + 2.

Ans: We have, $x^2$ = y and y = x + 2

         $\Rightarrow$ $x^2$ = x +2

         $\Rightarrow$ $x^2$ -( x +2) = 0 

          $\Rightarrow$ $x^2$ -x -2 = 0

          $\Rightarrow$ $x^2$ -2 x + x -2 = 0

         $\Rightarrow$ x (x - 2) + 1(x - 2) = 0

         $\Rightarrow$ (x +1) ( x - 2) = 0

        $\Rightarrow$  x = -1, 2

Therefore, area of shaded region A = ${\int }_{-1}^2$( x + 2  - $x^2$ ) dx 

 = ${[\frac{x^2}{2}+2x-\frac{x^3}{3}]}_{-1}^2$

= [$\frac{4}{2}+4-\frac{8}{3}-\frac{1}{2}$+ 2 -$\frac{1}{3}]$

= (8 - 3 - $\frac{1}{2}$) = $\frac{9}{2}$ sq units

7. Find the area of region bounded by the line x = 2 and the parabola $y^2$ = 8x .

Ans: We have, $y^2$ = 8x and x = 2.

Therefore, area of shaded region A = 2 ${\int }_0^2$($\sqrt{8x}$  ) dx

= 2. 2$\sqrt{2}$ ${\int }_0^2$($\sqrt{x}$  ) dx

= 4 .$\sqrt{2}$ ${\left[\frac{2x^{\frac{3}{2}}}{3}\right]}_0^2$

= 4 .$\sqrt{2}$ [$\frac{2{.2}^{\frac{3}{2}}}{3}$]

= $\frac{32}{3}$ sq units

8. Sketch the region {(x, 0) : y = $\sqrt{4-x^2}$} and x-axis. Find the area of the region using integration. 

Ans: Given region is {(x, 0) : y = $\sqrt{4-x^2}$} and x-axis

         We have, y = $\sqrt{4-x^2}$

          $\Rightarrow$ $y^2$ = 4 - $x^2$

      $\Rightarrow$ $y^2$ + $x^2$ = 4

Therefore, area of shaded region A = ${\int }_{-2}^2$($\sqrt{4-x^2}$ ) dx

 = ${\int }_{-2}^2$($\sqrt{2^2-x^2}$ ) dx

= ${[\frac{x}{2}\sqrt{2^2-x^2}+\frac{2^2}{2}si{n}^{-1}\frac{x}{2}]}_{-2}^2$

= $\frac{2}{2}$.0 + 2 . $\frac{\pi }{2}$+ $\frac{2}{2}.0$ - 2 $si{n}^{-1}$(-1)

= 2 . $\frac{\pi }{2}$ + 2 . $\frac{\pi }{2}$

= 2 $\pi$ sq units

9. Calculate the area under the curve y = 2 $\sqrt{x}$  included between the lines 

x = 0 and x = 1.

Ans: We have, y = 2 $\sqrt{x}$ , x =0 and  x = 1.

Therefore, area of shaded region A = ${\int }_0^1$(2$\sqrt{x}$ ) dx

= 2 ${\left[\frac{2x^{\frac{3}{2}}}{3}\right]}_0^1$ 

= 2 ($\frac{2}{3}.1$-0)

 = $\frac{4}{3}$ sq units

10. Using integration, find the area of the region bounded by the line 

2y = 5x + 7, x- axis and the lines x = 2 and x = 8.

Ans: We have 2 y = 5 x +7

        $\Rightarrow$ y= $\frac{5x}{2}+$ $\frac{7}{2}$

Area of shaded region = $\frac{1}{2}{\int }_2^8(5x+7)dx$

= $\frac{1}{2}{[\frac{5x^2}{2}+7x]}_2^8$

= $\frac{1}{2}$[$\frac{5{\left(8\right)}^2}{2}$ + 7(8) -$\frac{5{\left(2\right)}^2}{2}$-7(2)]

= 96 sq units

11. Draw a rough sketch of the curve y =$\sqrt{x-1}$ in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.

Ans: Given equation of the curve is y = $\sqrt{x-1}$ 

        $\Rightarrow y^2$ = x- 1

Therefore, area of shaded region A = ${\int }_1^5$($\sqrt{x-1}$ ) dx

=  ${\left[\frac{2{(x-1)}^{\frac{3}{2}}}{3}\right]}_1^5$ 

=  ($\frac{2}{3}.{(5-1)}^{\frac{3}{2}}$-0)

 = $\frac{16}{3}$ sq units

12. Determine the area under the curve y = $\sqrt{a^2-x^2}$ included between the lines x = 0 and x = a.

Ans: Given region is  y = $\sqrt{a^2-x^2}$  

          $\Rightarrow$ $y^2$ = $a^2$ - $x^2$

      $\Rightarrow$ $y^2$ + $x^2$ = $a^2$

Therefore, area of shaded region A = ${\int }_0^a$($\sqrt{a^2-x^2}$ ) dx

 = ${\int }_0^a$($\sqrt{a^2-x^2}$ ) dx

= ${[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{2}]}_0^a$

= [ 0 + $\frac{a^2}{2}si{n}^{-1}\frac{a}{2}$ -0 - $\frac{a^2}{2}si{n}^{-1}0$ ]

= $\frac{a^2}{2}$. $\frac{\pi }{2}$ 

= $\frac{a^2}{4}$  $\pi$ sq units

13. Find the area of the region bounded by y = $\sqrt{x}$ and y = x.

Ans: Given equations are y = $\sqrt{x}$ and y = x.

        $\Rightarrow$ x = $\sqrt{x}$  

        $\Rightarrow$ $x^2$ = x

       $\Rightarrow$ $x^2$ - x =0

       $\Rightarrow$ x (x - 1) = 0

       $\Rightarrow$ x = 0,1

Therefore, area of shaded region A = ${\int }_0^1$($\sqrt{x}$ ) dx - ${\int }_0^1$(x ) dx

=  ${\left[\frac{2{\left(x\right)}^{\frac{3}{2}}}{3}\right]}_0^1$ - ${\left[\frac{{\left(x\right)}^2}{2}\right]}_0^1$

=  ($\frac{2}{3}.1$- $\frac{1}{2}$)

 = $\frac{2}{3}$-$\frac{1}{2}$ 

= $\frac{1}{6}$ sq units

14. Find the area enclosed by the curve y = – $x^2$ and the straight line x + y + 2 = 0

Ans: We have ,y = – $x^2$ and the straight line x + y + 2 = 0.

  $\Rightarrow$ -x -2 = – $x^2$

   $\Rightarrow$  $x^2-x-2=0$

   $\Rightarrow$  $x^2+x-2x-2=0$

   $\Rightarrow$ x(x+1) - 2(x +1) =0 

   $\Rightarrow$ (x-2)(x +1) = 0

   $\Rightarrow$ x = 2, -1.

 Area of the shaded region = $|{\int }_{-1}^2(-x-2+x^2)dx|$

= $|{\int }_{-1}^2(x^2-x-2)dx|$

= ${\left|[\frac{{\left(x\right)}^3}{3}-\frac{{\left(x\right)}^2}{2}-2x]\right|}_{-1}^2$

= $\left|[\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2]\right|$ 

=$\left|\frac{16-12-24+2+3-12}{6}\right|=\left|\frac{-27}{6}\right|$

 $=\frac{9}{2}squnits$

15. Find the area bounded by the curve y = $\sqrt{x}$ , x = 2y + 3 in the first quadrant and x-  axis.

Ans: Given equation of the curves for y = $\sqrt{x}$ , x = 2y + 3 in the first quadrant.

On solving both the equations for y, we get

y = $\sqrt{2y+3}$

$\Rightarrow$ $y^2=2y+3$

$\Rightarrow$ $y^2-2y-3=0$

$\Rightarrow$ $y^2-3y+y-3=0$

$\Rightarrow$ y (y-3) + 1(y-3) =0

$\Rightarrow$ (y +1) (y-3) =0

$\Rightarrow$ y = -1,3

Area of shaded region = ${\int }_0^3(2y+3-y^2)dy$

= ${[\frac{2y^2}{2}+3y-\frac{y^3}{3}]}_0^3$

= [$\frac{18}{2}$ + (9) -9- 0]

= 9 sq units

16. Find the area of the region bounded by the curve  $y^2=2x$ and $x^2+y^2=4x$. 

Ans: We have $y^2=2x$ and $x^2+y^2=4x$

$\Rightarrow x^2+2x=4x$

$\Rightarrow x^2-2x=0$

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0$ Or $x=2$

Also, $x^2+y^2=4x$

$\Rightarrow x^2-4x=-y^2$

$\Rightarrow x^2-4x+4=-y^2+4$

$\Rightarrow {(x-2)}^2-2^2=-y^2$

∴ Required area = $2{\int }_0^2[\sqrt{2^2-{(x-2)}^2}-\sqrt{2x}]dx$

$=2[{\{\frac{x-2}{2}.\sqrt{2^2-{(x-2)}^2}+\frac{2^2}{2}si{n}^{-1}\left(\frac{x-2}{2}\right)\}}_0^2-{\{\sqrt{2}.\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\}}_0^2]$

$=2[(0+0-1.0+2.\frac{\pi }{2})-\frac{2\sqrt{2}}{3}(2^{\frac{3}{2}}-0)]$

$=\frac{4\pi }{2}-\frac{8.2}{3}$

$=2\pi -\frac{16}{3}$

$=2(\pi -\frac{8}{3})sq.units$

17. Find the area bounded by the curve $y=sinx$ between $x=0$ and $x=2\pi$.

Ans: Area of the shaded region = ${\int }_0^{2\pi }\left(sinx\right)dx$ =${\int }_0^{\pi }\left(sinx\right)dx$ +  $|{\int }_{\pi }^{2\pi }\left(sinx\right)dx|$

         =  ${[-cosx]}_0^{\pi }+\left|{[-cosx]}_{\pi }^{2\pi }\right|$

         = - [cos $\pi$ - cos 0]+ $|-[cos2\pi -cos\pi ]|$

= -[-1-1]+$|-(1+1)|$

= 2+2 = 4 sq units

18. Find the area of the region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.

Ans: Let we have the vertices of a $\mathrm{\Delta }ABC$ as A(–1, 1), B(0, 5) and C(3, 2).

Equation of AB is $y-1=\left(\frac{5-1}{0+1}\right)(x+1)$

$\Rightarrow y-1=4x+4$

$\Rightarrow y=4x+5---\left(i\right)$

And equation of BC is $y-5=\left(\frac{2-5}{3-0}\right)(x-0)$

$\Rightarrow y-5=\frac{-3}{3}\left(x\right)$

$\Rightarrow y=5-x---\left(ii\right)$

Similarly, equation of AC is $y-1=\left(\frac{2-1}{3+1}\right)(x+1)$

$\Rightarrow y-1=\frac{1}{4}(x+1)$

$\Rightarrow 4y=x+5---\left(iii\right)$

Area of shaded region $={\int }_{-1}^0(y_1-y_2)dx+{\int }_0^3(y_1-y_2)dx$

$={\int }_{-1}^0(4x+5-\frac{x+5}{4})dx+{\int }_0^3(5-x-\frac{x+5}{4})dx$

$={[\frac{4x^2}{2}+5x-\frac{x^2}{8}-\frac{5x}{4}]}_{-1}^0+{[5x-\frac{x^2}{2}-\frac{x^2}{8}-\frac{5x}{4}]}_0^3$

$=[0-(\frac{4{(-1)}^2}{2}+5(-1)-\frac{{(-1)}^2}{8}-\frac{5(-1)}{4})]+[5\left(3\right)-\frac{3^2}{2}-\frac{3^2}{8}-\frac{5\left(3\right)}{4}-0]$

$=[0-(2-5-\frac{1}{8}+\frac{5}{4})]+[15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}]$

$=3+\frac{1}{8}-\frac{5}{4}+15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}$

$=18+\left(\frac{1-10-36-9-30}{8}\right)$