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

The integral from-1 to 1 of x(1+x)^4 is 64/15

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integral from-1 to 1 of x(1+x)^4

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

\frac{64}{15}

Decimal

4.26666 \dots

Solution steps

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

Expand

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

= \int_{- 1}^{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

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

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

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

\int_{- 1}^{1} 4 x^{4} d x = \frac{8}{5}

\int_{- 1}^{1} x^{5} d x = 0

= 0 + \frac{8}{3} + 0 + \frac{8}{5} + 0

Simplify

0 + \frac{8}{3} + 0 + \frac{8}{5} + 0 : \frac{64}{15}

= \frac{64}{15}

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What is the integral from-1 to 1 of x(1+x)^4 ?
The integral from-1 to 1 of x(1+x)^4 is 64/15