What is the integral from 0 to 1 of x^5sqrt(1-x^6) ?

The integral from 0 to 1 of x^5sqrt(1-x^6) is 1/9

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integral from 0 to 1 of x^5sqrt(1-x^6)

\int_{0}^{1} x^{5} 1 - x^{6} d x

\frac{1}{9}

Decimal Notation

0.11111 \dots

Solution steps

\int_{0}^{1} x^{5} 1 - x^{6} d x

Apply u-substitution

: \int_{1}^{0} - \frac{u}{6} d u

= \int_{1}^{0} - \frac{u}{6} d u

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

= - \int_{0}^{1} - \frac{u}{6} d u

Take the constant out:

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

= - (- \frac{1}{6} \cdot \int_{0}^{1} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

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

Simplify

= \frac{1}{6} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{1}{6} \cdot \frac{2}{3}

Simplify

= \frac{1}{9}

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What is the integral from 0 to 1 of x^5sqrt(1-x^6) ?
The integral from 0 to 1 of x^5sqrt(1-x^6) is 1/9