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

The integral from 0 to 1 of (x(x^2-1)^3) is -1/8

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integral from 0 to 1 of (x(x^2-1)^3)

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

- \frac{1}{8}

Decimal Notation

- 0.125

Solution steps

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

Expand

x (x^{2} - 1)^{3} : x^{7} - 3 x^{5} + 3 x^{3} - x

= \int_{0}^{1} x^{7} - 3 x^{5} + 3 x^{3} - 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

= \int_{0}^{1} x^{7} d x - \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} 3 x^{3} d x - \int_{0}^{1} x d x

\int_{0}^{1} x^{7} d x = \frac{1}{8}

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

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

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

= \frac{1}{8} - \frac{1}{2} + \frac{3}{4} - \frac{1}{2}

Simplify

\frac{1}{8} - \frac{1}{2} + \frac{3}{4} - \frac{1}{2} : - \frac{1}{8}

= - \frac{1}{8}

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What is the integral from 0 to 1 of (x(x^2-1)^3) ?
The integral from 0 to 1 of (x(x^2-1)^3) is -1/8