The area of the region lying in the first quadrant by $y=4 x^2, x=0, y=2, y=4$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =\int_2^4 \frac{\sqrt{y}}{2} \mathrm{~d} y \ & =\frac{1}{2} \frac{\left[y^{\frac{3}{2}}ight]_2^4}{\frac{3}{2}} \ & =\frac{1}{3}(8-2 \sqrt{2}) 	ext { sq. units }\end{aligned}$

The area of the region lying in the first quadrant by $y=4 x^2, x=0, y=2, y=4$ is

Solution