What is the integral from 0 to 1 of x(2(1-x)) ?

The integral from 0 to 1 of x(2(1-x)) is 1/3

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integral from 0 to 1 of x(2(1-x))

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Take the constant out:

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

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

Expand

x (1 - x) : x - x^{2}

= 2 \cdot \int_{0}^{1} x - x^{2} 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

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

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

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

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

Simplify

2 (\frac{1}{2} - \frac{1}{3}) : \frac{1}{3}

= \frac{1}{3}

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What is the integral from 0 to 1 of x(2(1-x)) ?
The integral from 0 to 1 of x(2(1-x)) is 1/3