If $f(x)=x^{2 / 3}, x \geq 0 .$ Then, the area of the region enclosed by the curve $y=f(x)$ and the three lines $y=x, x=1$ and $x=8$ is

• A

  $\frac{63}{2}$
• B

  $\frac{93}{5}$
• C

  $\frac{105}{7}$
• Correct answer
  D

  $\frac{129}{10}$

Solution

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

Hence, the correct answer is option: 'D'.