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

The integral from-3 to-2 of 2x(4-x^2)^3 is 625/4

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integral from-3 to-2 of 2x(4-x^2)^3

\int_{- 3}^{- 2} 2 x (4 - x^{2})^{3} d x

\frac{625}{4}

Decimal Notation

156.25

Solution steps

\int_{- 3}^{- 2} 2 x (4 - 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 (4 - x^{2})^{3} d x

Apply u-substitution

: \int_{- 5}^{0} - \frac{u ^{3}}{2} d u

= 2 \cdot \int_{- 5}^{0} - \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_{- 5}^{0} u^{3} d u)

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{- 5}^{0}

= 2 (- \frac{1}{2} [\frac{u ^{4}}{4}]_{- 5}^{0})

Simplify

2 (- \frac{1}{2} [\frac{u ^{4}}{4}]_{- 5}^{0}) : - [\frac{u ^{4}}{4}]_{- 5}^{0}

= - [\frac{u ^{4}}{4}]_{- 5}^{0}

Compute the boundaries:

- \frac{625}{4}

= - (- \frac{625}{4})

Simplify

= \frac{625}{4}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{1 + x ^{2}})

\int \frac{1}{( 3 - x ) 2 - x} d x

\frac{\partial}{\partial y} (x^{2} + x y + y^{2} + y)

f (x) = (3 x^{3} + 1)^{2}

f (x) = sin (x)

What is the integral from-3 to-2 of 2x(4-x^2)^3 ?
The integral from-3 to-2 of 2x(4-x^2)^3 is 625/4