Question

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

Answer 1

Given curve $x^2=4y$ & $x=4y-2$

Area $={\int }_{-1}^2\left(\frac{x+2}{4}\right)-\frac{x^2}{4}dx$

$=\frac{1}{4}{[\frac{x^2}{2}+2x-\frac{x^3}{3}]}_{-1}^2$

$=\frac{1}{4}[(\frac{2^2}{2}+4-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3})]$

$=\frac{1}{4}[\frac{10}{3}+\frac{7}{6}]$

$=\frac{1}{4}\times \frac{27}{6}$

$=\frac{27}{24}$