What is the integral from-1 to 0 of-5x(x^2-1)^4 ?

The integral from-1 to 0 of-5x(x^2-1)^4 is 1/2

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integral from-1 to 0 of-5x(x^2-1)^4

\int_{- 1}^{0} - 5 x (x^{2} - 1)^{4} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{- 1}^{0} - 5 x (x^{2} - 1)^{4} d x

Take the constant out:

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

= - 5 \cdot \int_{- 1}^{0} x (x^{2} - 1)^{4} d x

Apply u-substitution

: \int_{0}^{- 1} \frac{u ^{4}}{2} d u

= - 5 \cdot \int_{0}^{- 1} \frac{u ^{4}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - 5 (- \int_{- 1}^{0} \frac{u ^{4}}{2} d u)

Take the constant out:

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

= - 5 (- \frac{1}{2} \cdot \int_{- 1}^{0} u^{4} d u)

Apply the Power Rule

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

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

Simplify

- 5 (- \frac{1}{2} [\frac{u ^{5}}{5}]_{- 1}^{0}) : \frac{5}{2} [\frac{u ^{5}}{5}]_{- 1}^{0}

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

Compute the boundaries:

\frac{1}{5}

= \frac{5}{2} \cdot \frac{1}{5}

Simplify

= \frac{1}{2}

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What is the integral from-1 to 0 of-5x(x^2-1)^4 ?
The integral from-1 to 0 of-5x(x^2-1)^4 is 1/2