What is the integral from 0 to 1 of x(1-x)^{193} ?

The integral from 0 to 1 of x(1-x)^{193} is 1/37830

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integral from 0 to 1 of x(1-x)^{193}

\int_{0}^{1} x (1 - x)^{193} d x

\frac{1}{37830}

Decimal Notation

0.00002 \dots

Solution steps

\int_{0}^{1} x (1 - x)^{193} d x

Apply u-substitution

: \int_{1}^{0} - (- u + 1) u^{193} d u

= \int_{1}^{0} - (- u + 1) u^{193} d u

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

= - \int_{0}^{1} - (- u + 1) u^{193} 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^{193} d u)

Expand

(- u + 1) u^{193} : - u^{194} + u^{193}

= - (- \int_{0}^{1} - u^{194} + u^{193} 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^{194} d u + \int_{0}^{1} u^{193} d u))

\int_{0}^{1} u^{194} d u = \frac{1}{195}

\int_{0}^{1} u^{193} d u = \frac{1}{194}

= - (- (- \frac{1}{195} + \frac{1}{194}))

Simplify

- (- (- \frac{1}{195} + \frac{1}{194})) : \frac{1}{37830}

= \frac{1}{37830}

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What is the integral from 0 to 1 of x(1-x)^{193} ?
The integral from 0 to 1 of x(1-x)^{193} is 1/37830