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

The integral from 0 to 1/3 of xsqrt(1-8x^2) is 13/324

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

\int_{0}^{\frac{1}{3}} x 1 - 8 x^{2} d x

\frac{13}{324}

Decimal Notation

0.04012 \dots

Solution steps

\int_{0}^{\frac{1}{3}} x 1 - 8 x^{2} d x

Apply u-substitution

: \int_{1}^{\frac{1}{9}} - \frac{u}{16} d u

= \int_{1}^{\frac{1}{9}} - \frac{u}{16} d u

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

= - \int_{\frac{1}{9}}^{1} - \frac{u}{16} d u

Take the constant out:

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

= - (- \frac{1}{16} \cdot \int_{\frac{1}{9}}^{1} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{1}{9}}^{1}

= - (- \frac{1}{16} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{1}{9}}^{1})

Simplify

= \frac{1}{16} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{1}{9}}^{1}

Compute the boundaries:

\frac{52}{81}

= \frac{1}{16} \cdot \frac{52}{81}

Simplify

= \frac{13}{324}

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What is the integral from 0 to 1/3 of xsqrt(1-8x^2) ?
The integral from 0 to 1/3 of xsqrt(1-8x^2) is 13/324