What is the integral from 0 to 3 of x^3sqrt(81-x^4) ?

The integral from 0 to 3 of x^3sqrt(81-x^4) is 243/2

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integral from 0 to 3 of x^3sqrt(81-x^4)

\int_{0}^{3} x^{3} 81 - x^{4} d x

\frac{243}{2}

Decimal

121.5

Solution steps

\int_{0}^{3} x^{3} 81 - x^{4} d x

Apply u-substitution

: \int_{81}^{0} - \frac{u}{4} d u

= \int_{81}^{0} - \frac{u}{4} d u

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

= - \int_{0}^{81} - \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}^{81} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{81}

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{81})

Simplify

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{81}

Compute the boundaries:

486

= \frac{1}{4} \cdot 486

Simplify

= \frac{243}{2}

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What is the integral from 0 to 3 of x^3sqrt(81-x^4) ?
The integral from 0 to 3 of x^3sqrt(81-x^4) is 243/2