evaluate the definite integral.\n int_{0}^{1} x^{2}(2 + 2x^{3})^{3} dx

Explanation:

Step1: Use substitution

Let $u = 2+2x^{3}$. Then $du=6x^{2}dx$, and $x^{2}dx=\frac{1}{6}du$. When $x = 0$, $u=2 + 2(0)^{3}=2$. When $x = 1$, $u=2+2(1)^{3}=4$.

Step2: Rewrite the integral

The integral $\int_{0}^{1}x^{2}(2 + 2x^{3})^{3}dx$ becomes $\frac{1}{6}\int_{2}^{4}u^{3}du$.

Step3: Integrate $u^{3}$

The antiderivative of $u^{3}$ is $\frac{1}{4}u^{4}$. So, $\frac{1}{6}\int_{2}^{4}u^{3}du=\frac{1}{6}\left[\frac{1}{4}u^{4}\right]_{2}^{4}$.

Step4: Evaluate the definite - integral

$\frac{1}{6}\times\frac{1}{4}(u^{4})\big|_{2}^{4}=\frac{1}{24}(4^{4}-2^{4})=\frac{1}{24}(256 - 16)=\frac{240}{24}=10$.

Answer:

$10$