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

The integral from 0 to 1 of x(2(x+1))/3 is 5/9

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integral from 0 to 1 of x(2(x+1))/3

\int_{0}^{1} x \frac{2 ( x + 1 )}{3} d x

\frac{5}{9}

Decimal Notation

0.55555 \dots

Solution steps

\int_{0}^{1} x \frac{2 ( x + 1 )}{3} d x

Take the constant out:

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

= \frac{2}{3} \cdot \int_{0}^{1} x (x + 1) d x

Expand

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

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

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

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

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

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

Simplify

\frac{2}{3} (\frac{1}{3} + \frac{1}{2}) : \frac{5}{9}

= \frac{5}{9}

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What is the integral from 0 to 1 of x(2(x+1))/3 ?
The integral from 0 to 1 of x(2(x+1))/3 is 5/9