Mathematics - Area Under Curves Question with Solution | TestHub

Given, $f(x)=x^{2/3},x\ge 0$ and line $y=x$$\therefore$ Required area $A={\int }_{x=1}^8\left(x-x^{2/3}\right)dx$ $={\left[\frac{x^2}{2}-\frac{3}{5}{x}^{5/3}\right]}_1^9=\left(32-\frac{3}{5}\times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)$ $=32\times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}$ $=\frac{128+1}{10}=\frac{129}{10}$