Question

find the area bounded by the curve y2+x-4y=5 and the y axis?​

Answer 1

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

Given curve is

$\rm :\longmapsto\: {y}^{2} + x - 4y = 5$

Let first we convert in to standard form

$\rm :\longmapsto\: {y}^{2} - 4y = 5 - x$

On adding 4 both sides, we get

$\rm :\longmapsto\: {y}^{2} - 4y + 4= 5 - x + 4$

$\rm :\longmapsto\: {(y - 2)}^{2} = - (x - 9) - - (1)$

Its represents the left handed parabola whose vertex us ( 9, 2 ).

Let find the point of intersection with y - axis,

On y - axis, x = 0

So, equation (1), becomes

$\rm :\longmapsto\: {(y - 2)}^{2} = - (0 - 9)$

$\rm :\longmapsto\: {(y - 2)}^{2} =9$

$\rm :\longmapsto\:y - 2\: = \: \pm \: 3$

$\bf\implies \:y = 5 \: \: or \: \: y = - 1$

Hence, the point of intersection of the curve with y - axis is (0, 5) and (0, - 1).

So, the required area bounded by the curve with y - axis is given by

$\rm :\longmapsto\:Area \: = \displaystyle\int_{ - 1}^5 \tt \:x \: dy$

$\rm :\longmapsto\:Area \: = \displaystyle\int_{ - 1}^5 \tt \: \bigg(9 - {(y - 2)}^{2} \bigg) \: dy$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:using \: (1) \bigg \}}$

$\rm \: \: = \: \bigg(9y - \dfrac{ {(y - 2)}^{3} }{3} \bigg)_{ - 1}^5$

$\rm \: \: = \: \bigg(45 - \dfrac{ {(5 - 2)}^{3} }{3} \bigg) - \bigg( - 9 - \dfrac{ {( - 1 - 2)}^{3} }{3} \bigg)$

$\rm \: \: = \: \bigg(45 - \dfrac{ {(3)}^{3} }{3} \bigg) - \bigg( - 9 - \dfrac{ {( -3)}^{3} }{3} \bigg)$

$\rm \: \: =45 - 9 + 9 - 9$

$\rm \: \: = \: 36 \: square \: units$

Answer 2

Answer:

The Answer will be 36 after solve the equation

Step-by-step explanation: