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

The integral from-3 to-2 of (2x(3-x^2)^3) is 1295/4

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

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

\frac{1295}{4}

Decimal

323.75

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 6}^{- 1} - \frac{u ^{3}}{2} d u

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

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 6}^{- 1}

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

Simplify

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

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

Compute the boundaries:

- \frac{1295}{4}

= - (- \frac{1295}{4})

Simplify

= \frac{1295}{4}

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\frac{d}{d x} (x^{2} arccos (x))

What is the integral from-3 to-2 of (2x(3-x^2)^3) ?
The integral from-3 to-2 of (2x(3-x^2)^3) is 1295/4