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

The integral from 0 to 1 of x(1-x)^{1/2} is 4/15

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

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

\frac{4}{15}

Decimal

0.26666 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - 2 (- u^{2} + 1) u^{2} d u

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

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

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

Take the constant out:

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

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

Expand

(- u^{2} + 1) u^{2} : - u^{4} + u^{2}

= - (- 2 \cdot \int_{0}^{1} - u^{4} + u^{2} 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

= - (- 2 (- \int_{0}^{1} u^{4} d u + \int_{0}^{1} u^{2} d u))

\int_{0}^{1} u^{4} d u = \frac{1}{5}

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

= - (- 2 (- \frac{1}{5} + \frac{1}{3}))

Simplify

- (- 2 (- \frac{1}{5} + \frac{1}{3})) : \frac{4}{15}

= \frac{4}{15}

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What is the integral from 0 to 1 of x(1-x)^{1/2} ?
The integral from 0 to 1 of x(1-x)^{1/2} is 4/15