Question

Please help! dont know how to go about solving this definite integral. -2 to 4 x^2(x^3+8)dx

Answer 1

$$\int_{-2}^{4}x^2(x^3+8)dx$$

$$u=x^3+8$$

$$du=3x^2dx$$

$$\frac{du}{3}=x^2dx$$

$$\frac{1}{3}\int_{0}^{72}udu$$

$$=\frac{1}{3}\left.\frac{u^2}{2}\right|_{0}^{72}=864$$