What is the integral from 0 to 1 of 5x(x^2-1)^5 ?

The integral from 0 to 1 of 5x(x^2-1)^5 is -5/12

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integral from 0 to 1 of 5x(x^2-1)^5

\int_{0}^{1} 5 x (x^{2} - 1)^{5} d x

- \frac{5}{12}

Decimal

- 0.41666 \dots

Solution steps

\int_{0}^{1} 5 x (x^{2} - 1)^{5} 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 (x^{2} - 1)^{5} d x

Apply u-substitution

: \int_{- 1}^{0} \frac{u ^{5}}{2} d u

= 5 \cdot \int_{- 1}^{0} \frac{u ^{5}}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int_{- 1}^{0} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 1}^{0}

= 5 \cdot \frac{1}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

Simplify

5 \cdot \frac{1}{2} [\frac{u ^{6}}{6}]_{- 1}^{0} : \frac{5}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

= \frac{5}{2} [\frac{u ^{6}}{6}]_{- 1}^{0}

Compute the boundaries:

- \frac{1}{6}

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

Simplify

= - \frac{5}{12}

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What is the integral from 0 to 1 of 5x(x^2-1)^5 ?
The integral from 0 to 1 of 5x(x^2-1)^5 is -5/12