What is the integral from-2 to-1 of 6x(2x^2-6)^3 ?

The integral from-2 to-1 of 6x(2x^2-6)^3 is 90

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integral from-2 to-1 of 6x(2x^2-6)^3

\int_{- 2}^{- 1} 6 x (2 x^{2} - 6)^{3} d x

90

Solution steps

\int_{- 2}^{- 1} 6 x (2 x^{2} - 6)^{3} d x

Take the constant out:

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

= 6 \cdot \int_{- 2}^{- 1} x (2 x^{2} - 6)^{3} d x

Apply u-substitution

: \int_{2}^{- 4} \frac{u ^{3}}{4} d u

= 6 \cdot \int_{2}^{- 4} \frac{u ^{3}}{4} d u

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

= 6 (- \int_{- 4}^{2} \frac{u ^{3}}{4} d u)

Take the constant out:

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

= 6 (- \frac{1}{4} \cdot \int_{- 4}^{2} u^{3} d u)

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 4}^{2}

= 6 (- \frac{1}{4} [\frac{u ^{4}}{4}]_{- 4}^{2})

Simplify

6 (- \frac{1}{4} [\frac{u ^{4}}{4}]_{- 4}^{2}) : - \frac{3}{2} [\frac{u ^{4}}{4}]_{- 4}^{2}

= - \frac{3}{2} [\frac{u ^{4}}{4}]_{- 4}^{2}

Compute the boundaries:

- 60

= - \frac{3}{2} (- 60)

Simplify

= 90

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What is the integral from-2 to-1 of 6x(2x^2-6)^3 ?
The integral from-2 to-1 of 6x(2x^2-6)^3 is 90