Mathematics - Area Under Curves Question with Solution | TestHub

Given equations of the curves are $x^2=4y$ and $x=4y-2$.

To find the points of intersection, substitute $4y=x+2$ into the equation of the parabola:

$x^2=x+2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

$x=-1,2$

The required area is bounded between $x=-1$ and $x=2$. In this interval, the line lies above the parabola.

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

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

Area $=\frac{1}{4}\left(\left(\frac{4}{2}+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)\right)$

Area $=\frac{1}{4}\left(\left(6-\frac{8}{3}\right)-\left(-\frac{7}{6}\right)\right)$

Area $=\frac{1}{4}\left(\frac{10}{3}+\frac{7}{6}\right)$

Area $=\frac{1}{4}\left(\frac{27}{6}\right)=\frac{9}{8}$ sq. units.

Answer: $\frac{9}{8}$