What is the integral from 0 to 1 of-6x(x^2-2)^4 ?

The integral from 0 to 1 of-6x(x^2-2)^4 is -93/5

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integral from 0 to 1 of-6x(x^2-2)^4

\int_{0}^{1} - 6 x (x^{2} - 2)^{4} d x

- \frac{93}{5}

Decimal

- 18.6

Solution steps

\int_{0}^{1} - 6 x (x^{2} - 2)^{4} d x

Take the constant out:

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

= - 6 \cdot \int_{0}^{1} x (x^{2} - 2)^{4} d x

Apply u-substitution

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

= - 6 \cdot \int_{- 2}^{- 1} \frac{u ^{4}}{2} d u

Take the constant out:

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

= - 6 \cdot \frac{1}{2} \cdot \int_{- 2}^{- 1} u^{4} d u

Apply the Power Rule

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

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

Simplify

- 6 \cdot \frac{1}{2} [\frac{u ^{5}}{5}]_{- 2}^{- 1} : - 3 [\frac{u ^{5}}{5}]_{- 2}^{- 1}

= - 3 [\frac{u ^{5}}{5}]_{- 2}^{- 1}

Compute the boundaries:

\frac{31}{5}

= - 3 \cdot \frac{31}{5}

Simplify

= - \frac{93}{5}

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What is the integral from 0 to 1 of-6x(x^2-2)^4 ?
The integral from 0 to 1 of-6x(x^2-2)^4 is -93/5