Question

Find the area bounded by the curves $x^2=4y$ and the straight line $x-4y+2=0$

Answer 1

$x^2=4yx-4y+2=0$

$x-x^2+2=0$

$x^2-x-2=0$

$x(x-2)+1(x-2)$

$x=-1,2$

$A={\int }_{-1}^2\frac{x+2}{4}-\frac{x^2}{4}dxA={[\frac{x^2}{8}+\frac{x}{2}-\frac{x^3}{12}]}_{-1}^{2}A=-[\frac{{-1}^2}{8}+\frac{-1}{2}-\frac{{-1}^3}{12}-(\frac{2^2}{8}+\frac{2}{2}-\frac{2^3}{12}\left)\right]A=-\frac{1}{8}+\frac{1}{2}-\frac{1}{12}+\frac{1}{2}+1-\frac{4}{3}A=\frac{10}{12}+\frac{7}{24}=\frac{9}{8}$