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

The integral from 0 to 1/2 of xsqrt(1-x^2) is 1/3-(sqrt(3))/8

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

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

\frac{1}{3} - \frac{3}{8}

Decimal

0.11682 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{\frac{3}{2}} - u^{2} d u

= \int_{1}^{\frac{3}{2}} - u^{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{3}{2}}^{1} - u^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{\frac{3}{2}}^{1} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1}

= - (- [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1})

Simplify

= [\frac{u ^{3}}{3}]_{\frac{3}{2}}^{1}

Compute the boundaries:

\frac{1}{3} - \frac{3}{8}

= \frac{1}{3} - \frac{3}{8}

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What is the integral from 0 to 1/2 of xsqrt(1-x^2) ?
The integral from 0 to 1/2 of xsqrt(1-x^2) is 1/3-(sqrt(3))/8