What is the integral from 0 to 1 of x^2sqrt(9-x^3) ?

The integral from 0 to 1 of x^2sqrt(9-x^3) is (54-32sqrt(2))/9

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integral from 0 to 1 of x^2sqrt(9-x^3)

\int_{0}^{1} x^{2} 9 - x^{3} d x

\frac{54 - 32 2}{9}

Decimal

0.97168 \dots

Solution steps

\int_{0}^{1} x^{2} 9 - x^{3} d x

Apply u-substitution

: \int_{9}^{8} - \frac{u}{3} d u

= \int_{9}^{8} - \frac{u}{3} d u

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

= - \int_{8}^{9} - \frac{u}{3} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{8}^{9} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{8}^{9}

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

Simplify

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

Compute the boundaries:

18 - \frac{32 2}{3}

= \frac{1}{3} (18 - \frac{32 2}{3})

Simplify

= \frac{54 - 32 2}{9}

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What is the integral from 0 to 1 of x^2sqrt(9-x^3) ?
The integral from 0 to 1 of x^2sqrt(9-x^3) is (54-32sqrt(2))/9