∫limits_0^8(x^2/3+1)dx is equal to

The correct answer is Option (4) → $\frac{136}{5}$

Given integral:

$\int_{0}^{8}(x^{\frac{2}{3}}+1)\,dx$

$=\int_{0}^{8}x^{\frac{2}{3}}\,dx+\int_{0}^{8}1\,dx$

$=\left[\frac{3}{5}x^{\frac{5}{3}}+x\right]_{0}^{8}$

$=\frac{3}{5}(8)^{\frac{5}{3}}+8$

Since $(8)^{\frac{1}{3}}=2\Rightarrow(8)^{\frac{5}{3}}=2^{5}=32$

$\Rightarrow\ \frac{3}{5}\times32+8=\frac{96}{5}+8=\frac{136}{5}$

Final Answer: $\frac{136}{5}$