Using integration, find the area bounded by the curve `x^2=4y` and the line `x=4y-2.`

Options

A. `3/8"sq units"`
B. `5/8" sq unit "`
C. `7/8 " sq unit"`
D. `9/8" sq units"`

Answer

D

Explanation

Given equation of curve is `x^(2)4y" and the straight line " x=4y-2`

For intresection point, put `x = 4y-2` in equation of curve, we get
`(4y-2)^(2)=4y`
`rArr 16y^(2)+4-16y=4y`
`rArr 16y^(2)-20y+4=0`
`rArr 4y^(2)-5y+1=0`
`rArr 4y(y-1)-1(y-1)=0`
`rArr (4y-1)(y-1)=0`
`:. y=1,1/4`
For `y=1, x=sqrt4.1=2` [since, negative value does not satisfy the equation of line]
For `y=1/4,x=sqrt4. 1/4=-1` [positive value does not satisfy the equation of line]
So, the intersection points are (2, 1) and `(-1, 1/4)`
`:. " Area of shaded region "=int_(-1)^(2)((x+2)/4)dx-int_(-1)^(2)(x^(2))/4dx`
`=1/4[x^(2)/2+2x]_(-1)^(2)-1/4abs(x^(3)/3)_(-1)^(2)`
`=-1/4[4/2+4--1/2+2]-1/4[8/3+1/3]`
`=1/4 . 15/2-1/4 . 9/3=(45-18)/24`
`=27/24=9/8" sq units"`