Question

Find the area of the region lying in the first quadrant and bounded by $y=4x^2,x=0,y=1$ and y = 4.

Answer 1

The area in the first quadrant bounded by $y=4x^2,y=4,x=0,y=1$ is represented by the shaded region.

$\therefore$ Required area (shown in shaded region)

${\int }_1^4|x|dy={\int }_1^4\frac{1}{2}\sqrt{y}dy(\because y=4x^2,\therefore x^2=\frac{y}{4}\Rightarrow |x|=\frac{\sqrt{y}}{2})=\frac{1}{2}[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}{]}_1^4=\frac{1}{3}(4^{\frac{3}{2}}-1^{\frac{3}{2}})=\frac{1}{3}[(2^2{)}^{\frac{3}{2}}-1]=\frac{1}{3}(8-1)=\frac{7}{8}squnit$