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

The integral from 0 to 1 of (x(-x^2+1)^{3/2})/3 is 1/15

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

\int_{0}^{1} \frac{x ( - x ^{2} + 1 ) ^{\frac{3}{2}}}{3} d x

\frac{1}{15}

Decimal

0.06666 \dots

Solution steps

\int_{0}^{1} \frac{x ( - x ^{2} + 1 ) ^{\frac{3}{2}}}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} x (- x^{2} + 1)^{\frac{3}{2}} d x

Apply u-substitution

: \int_{1}^{0} - \frac{u ^{\frac{3}{2}}}{2} d u

= \frac{1}{3} \cdot \int_{1}^{0} - \frac{u ^{\frac{3}{2}}}{2} d u

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

= \frac{1}{3} (- \int_{0}^{1} - \frac{u ^{\frac{3}{2}}}{2} d u)

Take the constant out:

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

= \frac{1}{3} (- (- \frac{1}{2} \cdot \int_{0}^{1} u^{\frac{3}{2}} d u))

Apply the Power Rule

: [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{1}

= \frac{1}{3} (- (- \frac{1}{2} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{1}))

Simplify

\frac{1}{3} (- (- \frac{1}{2} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{1})) : \frac{1}{6} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{1}

= \frac{1}{6} [\frac{2}{5} u^{\frac{5}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{5}

= \frac{1}{6} \cdot \frac{2}{5}

Simplify

= \frac{1}{15}

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What is the integral from 0 to 1 of (x(-x^2+1)^{3/2})/3 ?
The integral from 0 to 1 of (x(-x^2+1)^{3/2})/3 is 1/15