11. use a change of variables to evaluate the following integral: $int_{0}^{2}x^{3}sqrt{16 - x^{4}}dx$

Explanation:

Step1: Set substitution variable

Let $u = 16 - x^{4}$. Then $du=-4x^{3}dx$, and $x^{3}dx=-\frac{1}{4}du$.

Step2: Change limits of integration

When $x = 0$, $u=16 - 0^{4}=16$. When $x = 2$, $u=16 - 2^{4}=16 - 16 = 0$.

Step3: Rewrite the integral

The integral $\int_{0}^{2}x^{3}\sqrt{16 - x^{4}}dx$ becomes $-\frac{1}{4}\int_{16}^{0}\sqrt{u}du$.

Step4: Evaluate the integral

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $n=\frac{1}{2}$, $\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}+C$. So, $-\frac{1}{4}\int_{16}^{0}\sqrt{u}du=-\frac{1}{4}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{16}^{0}$.

Step5: Calculate the definite - integral value

$-\frac{1}{4}\times\frac{2}{3}(0 - 16^{\frac{3}{2}})=-\frac{1}{6}(0 - 64)=\frac{32}{3}$.

Answer:

$\frac{32}{3}$