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

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

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

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

\frac{243}{2}

Decimal Notation

121.5

Solution steps

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

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}

\int_{2 x^{2}}^{x^{4}} 2 t + t^{2} d t

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

\int_{0}^{1} x^{3} (2 + x^{4}) d x

\int_{- 2}^{0} - c x d x

\int_{0}^{π} e^{3 x} \cdot sin (x) d x

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