What is the integral from-2 to 0 of x^3sqrt(16-x^4) ?

The integral from-2 to 0 of x^3sqrt(16-x^4) is -32/3

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integral from-2 to 0 of x^3sqrt(16-x^4)

\int_{- 2}^{0} x^{3} 16 - x^{4} d x

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

\int_{- 2}^{0} x^{3} 16 - x^{4} d x

Apply u-substitution

: \int_{0}^{16} - \frac{u}{4} d u

= \int_{0}^{16} - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{0}^{16} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

= - \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{16}

Compute the boundaries:

\frac{128}{3}

= - \frac{1}{4} \cdot \frac{128}{3}

Simplify

= - \frac{32}{3}

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What is the integral from-2 to 0 of x^3sqrt(16-x^4) ?
The integral from-2 to 0 of x^3sqrt(16-x^4) is -32/3