What is the integral from 0 to 1 of 5x(1-x)^4 ?

The integral from 0 to 1 of 5x(1-x)^4 is 1/6

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integral from 0 to 1 of 5x(1-x)^4

\int_{0}^{1} 5 x (1 - x)^{4} d x

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

\int_{0}^{1} 5 x (1 - x)^{4} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} x (1 - x)^{4} d x

Expand

x (1 - x)^{4} : x - 4 x^{2} + 6 x^{3} - 4 x^{4} + x^{5}

= 5 \cdot \int_{0}^{1} x - 4 x^{2} + 6 x^{3} - 4 x^{4} + x^{5} d x

Apply the Sum Rule:

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

= 5 (\int_{0}^{1} x d x - \int_{0}^{1} 4 x^{2} d x + \int_{0}^{1} 6 x^{3} d x - \int_{0}^{1} 4 x^{4} d x + \int_{0}^{1} x^{5} d x)

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 4 x^{2} d x = \frac{4}{3}

\int_{0}^{1} 6 x^{3} d x = \frac{3}{2}

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

\int_{0}^{1} x^{5} d x = \frac{1}{6}

= 5 (\frac{1}{2} - \frac{4}{3} + \frac{3}{2} - \frac{4}{5} + \frac{1}{6})

Simplify

5 (\frac{1}{2} - \frac{4}{3} + \frac{3}{2} - \frac{4}{5} + \frac{1}{6}) : \frac{1}{6}

= \frac{1}{6}

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What is the integral from 0 to 1 of 5x(1-x)^4 ?
The integral from 0 to 1 of 5x(1-x)^4 is 1/6