# 5 $\\int_{0}^{1}x\\sqrt{1 - x^{2}}dx$

Explanation:

Step1: Substitute $u = 1 - x^{2}$

Differentiate $u$ with respect to $x$: $du=-2xdx$, so $xdx=-\frac{1}{2}du$. When $x = 0$, $u=1 - 0^{2}=1$; when $x = 1$, $u=1 - 1^{2}=0$. The integral $\int_{0}^{1}x\sqrt{1 - x^{2}}dx$ becomes $-\frac{1}{2}\int_{1}^{0}\sqrt{u}du$.

Step2: Use the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$. For $\int\sqrt{u}du=\int u^{\frac{1}{2}}du=\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{2}{3}u^{\frac{3}{2}}+C$. Now, $-\frac{1}{2}\int_{1}^{0}\sqrt{u}du=-\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}$.

Step3: Evaluate the definite integral

First, substitute the upper and lower limits: $-\frac{1}{2}\left(\frac{2}{3}(0)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}\right)$. Simplify the expression: $-\frac{1}{2}\times\left(-\frac{2}{3}\right)$.

Answer:

$\frac{1}{3}$