What is the integral from 0 to 4 of x^3sqrt(256-x^4) ?

The integral from 0 to 4 of x^3sqrt(256-x^4) is 2048/3

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integral from 0 to 4 of x^3sqrt(256-x^4)

\int_{0}^{4} x^{3} 256 - x^{4} d x

\frac{2048}{3}

Decimal

682.66666 \dots

Solution steps

\int_{0}^{4} x^{3} 256 - x^{4} d x

Apply u-substitution

: \int_{256}^{0} - \frac{u}{4} d u

= \int_{256}^{0} - \frac{u}{4} d u

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

= - \int_{0}^{256} - \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}^{256} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{256}

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{256})

Simplify

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{256}

Compute the boundaries:

\frac{8192}{3}

= \frac{1}{4} \cdot \frac{8192}{3}

Simplify

= \frac{2048}{3}

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What is the integral from 0 to 4 of x^3sqrt(256-x^4) ?
The integral from 0 to 4 of x^3sqrt(256-x^4) is 2048/3