Question

The area above x-axis, bounded by the line $x=4$ and the curve $y=f\left(x\right)$, where $f\left(x\right)=x^2,0\le x\le 1$ and $f\left(x\right)=\sqrt{x},x\ge 1$, is

Answer 1

The required area
$={\int }_0^1{x}^{2}dx+{\int }_1^4\sqrt{x}dx={\left[\frac{x^3}{3}\right]}_0^1+{\left[\frac{2x^{3/2}}{3}\right]}_1^4$
$=\frac{1}{3}+\frac{2}{3}(8-1)=5$