What is the integral from 0 to 3/4 of xsqrt(9-4x^2) ?

The integral from 0 to 3/4 of xsqrt(9-4x^2) is (72-27sqrt(3))/(32)

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integral from 0 to 3/4 of xsqrt(9-4x^2)

\int_{0}^{\frac{3}{4}} x 9 - 4 x^{2} d x

\frac{72 - 27 3}{32}

Decimal Notation

0.78858 \dots

Solution steps

\int_{0}^{\frac{3}{4}} x 9 - 4 x^{2} d x

Apply u-substitution

: \int_{9}^{\frac{27}{4}} - \frac{u}{8} d u

= \int_{9}^{\frac{27}{4}} - \frac{u}{8} d u

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

= - \int_{\frac{27}{4}}^{9} - \frac{u}{8} d u

Take the constant out:

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

= - (- \frac{1}{8} \cdot \int_{\frac{27}{4}}^{9} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{27}{4}}^{9}

= - (- \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{27}{4}}^{9})

Simplify

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{27}{4}}^{9}

Compute the boundaries:

18 - \frac{27 3}{4}

= \frac{1}{8} (18 - \frac{27 3}{4})

Simplify

= \frac{72 - 27 3}{32}

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What is the integral from 0 to 3/4 of xsqrt(9-4x^2) ?
The integral from 0 to 3/4 of xsqrt(9-4x^2) is (72-27sqrt(3))/(32)