What is the integral from 0 to 1 of (1-x)^{30}x ?

The integral from 0 to 1 of (1-x)^{30}x is 1/992

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integral from 0 to 1 of (1-x)^{30}x

\int_{0}^{1} (1 - x)^{30} x d x

\frac{1}{992}

Decimal

0.00100 \dots

Solution steps

\int_{0}^{1} (1 - x)^{30} x d x

Apply u-substitution

: \int_{1}^{0} - (- u + 1) u^{30} d u

= \int_{1}^{0} - (- u + 1) u^{30} d u

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

= - \int_{0}^{1} - (- u + 1) u^{30} d u

Take the constant out:

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

= - (- \int_{0}^{1} (- u + 1) u^{30} d u)

Expand

(- u + 1) u^{30} : - u^{31} + u^{30}

= - (- \int_{0}^{1} - u^{31} + u^{30} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{1} u^{31} d u + \int_{0}^{1} u^{30} d u))

\int_{0}^{1} u^{31} d u = \frac{1}{32}

\int_{0}^{1} u^{30} d u = \frac{1}{31}

= - (- (- \frac{1}{32} + \frac{1}{31}))

Simplify

- (- (- \frac{1}{32} + \frac{1}{31})) : \frac{1}{992}

= \frac{1}{992}

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What is the integral from 0 to 1 of (1-x)^{30}x ?
The integral from 0 to 1 of (1-x)^{30}x is 1/992