The area (in sq. units) of the region bounded by the curve $x^2=4 y$ and the straight line $x=4 y-2$ is

Question

Aicharya · Accepted Answer

The points of intersection of $x^2=4 y$ and $x=4 y-2$ are $(2,1)$ and $\left(-1, \frac{1}{4}\right)$.

Required area $\begin{aligned} & =\int_{-1}^2 \frac{1}{4}(x+2) \mathrm{d} x-\int_{-1}^2 \frac{1}{4} x^2 \mathrm{~d} x \\ & =\frac{1}{4}\left[\frac{x^2}{2}+2 x\right]_{-1}^2-\frac{1}{4}\left[\frac{x^3}{3}\right]_{-1}^2 \\ & =\frac{9}{8} \text { sq. units } \end{aligned}$

The area (in sq. units) of the region bounded by the curve $x^2=4 y$ and the straight line $x=4 y-2$ is

Solution