Question

Area of the region bounded by y^2=4xandy=2x-4​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curves are

$\rm :\longmapsto\: {y}^{2} = 4x - - - (1)$

and

$\rm :\longmapsto\:y = 2x - 4 - - - (2)$

Point of intersection of curve (1) and (2)

$\rm :\longmapsto\: {y}^{2} = 4x$

$\rm :\longmapsto\: {y}^{2} = 2 \times 2x$

$\rm :\longmapsto\: {y}^{2} = 2 \times (y + 4)$

$\rm :\longmapsto\: {y}^{2} = 2y +8$

$\rm :\longmapsto\: {y}^{2} - 2y - 8 = 0$

$\rm :\longmapsto\: {y}^{2} - 4y + 2y - 8 = 0$

$\rm :\longmapsto\:y(y - 4) + 2(y - 4) = 0$

$\rm :\longmapsto\:(y - 4)(y + 2) = 0$

$\rm :\longmapsto\:y = 4 \: \: \: or \: \: \: y = - \: 2$

Hᴇɴᴄᴇ,

➢ Points of intersection of the given curve are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 4 & \sf 4 \\ \\ \sf 1 & \sf - 2 \end{array}} \\ \end{gathered}$

Curve sketching,

See the attachment

Required Area with respect to y - axis

$\rm  \:  =  \: \:\displaystyle\int_{ - 2}^4\sf \: (x_{line} - x_{parabola}) \: dy$

$\rm  \:  =  \: \:\displaystyle\int_{ - 2}^4\sf \: ( \dfrac{y + 4}{2} - \dfrac{ {y}^{2} }{4} )\: dy$

$\rm  \:  =  \: \dfrac{1}{2} \:\displaystyle\int_{ - 2}^4\sf \: ( y + 4- \dfrac{ {y}^{2} }{2} )\: dy$

$\rm  \:  =  \: \:\dfrac{1}{2}\bigg(\dfrac{ {y}^{2} }{2} + 4y - \dfrac{ {y}^{3} }{6} \bigg)_{ - 2}^4$

$\rm  \:  =  \: \:\dfrac{1}{2}\bigg(8 + 16 - \dfrac{32}{3} - 2 + 8 - \dfrac{4}{3} \bigg)$

$\rm  \:  =  \: \:\dfrac{1}{2}\bigg(30 - 12 \bigg)$

$\rm  \:  =  \: \:\dfrac{1}{2}\bigg(18 \bigg)$

$\rm  \:  =  \: \:9 \: square \: units$

Formula Used :-

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: {x}^{n} \: dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: k \: dx \: = \: kx \: + \: c}}$