What is the integral from 0 to 1 of x(1-x^2)^{1/2} ?

The integral from 0 to 1 of x(1-x^2)^{1/2} is 1/3

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integral from 0 to 1 of x(1-x^2)^{1/2}

\int_{0}^{1} x (1 - x^{2})^{\frac{1}{2}} d x

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

\int_{0}^{1} x (1 - x^{2})^{\frac{1}{2}} d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} - \int - \frac{x ^{3}}{2 ( 1 - x ^{2} ) ^{\frac{1}{2}}} d x]_{0}^{1}

= [\frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} - \int - \frac{x ^{3}}{2 ( 1 - x ^{2} ) ^{\frac{1}{2}}} d x]_{0}^{1}

\int - \frac{x ^{3}}{2 ( 1 - x ^{2} ) ^{\frac{1}{2}}} d x = - \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2})

= [\frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} - (- \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2}))]_{0}^{1}

Simplify

= [\frac{1}{2} x^{2} (1 - x^{2})^{\frac{1}{2}} + \frac{1}{6} ((1 - x^{2})^{\frac{3}{2}} - 3 1 - x^{2})]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3}

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What is the integral from 0 to 1 of x(1-x^2)^{1/2} ?
The integral from 0 to 1 of x(1-x^2)^{1/2} is 1/3