use a change of variables or the table to evaluate the following definite integral. $int_{0}^{1}x^{3}sqrt{1…

Explanation:

Step1: Use substitution

Let $u = 1 - x^{4}$, then $du=-4x^{3}dx$, and $x^{3}dx=-\frac{1}{4}du$. When $x = 0$, $u=1 - 0^{4}=1$; when $x = 1$, $u=1 - 1^{4}=0$.

Step2: Rewrite the integral

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

Step3: Evaluate the integral

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

Step4: Calculate the definite - integral value

$-\frac{1}{4}\left(\frac{2}{3}\times0^{\frac{3}{2}}-\frac{2}{3}\times1^{\frac{3}{2}}\right)=-\frac{1}{4}\times\left(0 - \frac{2}{3}\right)=\frac{1}{6}$.

Answer:

$\frac{1}{6}$