Question

Find the area between curve $x^2=4y$ and the line $x=4y-2$.

Answer 1

$x^2=4y\Rightarrow y=\frac{x^2}{4}....\left(1\right)$

$x=4y-2$

$0=4y-x-\mathrm{2....}\left(2\right)$

From equation (1) and (2)

$0=\frac{{4x}^2}{4}-x-2$

$0=x^2-x-2$

$y=\frac{1}{4},y=1$

Area bounded between curve and line

$=$ Area of $ALMB-$ Area of $ALO+$ Area of $OMB$)

$={\int }_{-1}^2\frac{x}{4}+\frac{1}{2}dx-\left[{\int }_{-1}^0\frac{x^2}{4}dx+{\int }_0^2\frac{x^2}{4}dx\right]$

$={[\frac{x^2}{4}+\frac{x}{2}]}_{-1}^2-[{\left[\frac{x^3}{12}\right]}_{-1}^0+{\left[\frac{x^3}{12}\right]}_0^2]$

$=\frac{4}{8}+1-\frac{1}{8}+\frac{1}{2}-\frac{1}{12}-\frac{8}{12}=\frac{3}{8}+\frac{3}{2}-\frac{3}{4}=\frac{9}{8}$