What is the integral from 0 to 3 of x^3sqrt(9-x^2) ?

The integral from 0 to 3 of x^3sqrt(9-x^2) is 162/5

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integral from 0 to 3 of x^3sqrt(9-x^2)

\int_{0}^{3} x^{3} 9 - x^{2} d x

\frac{162}{5}

Decimal Notation

32.4

Solution steps

\int_{0}^{3} x^{3} 9 - x^{2} d x

Apply u-substitution

: \int_{3}^{0} - (- u^{2} + 9) u^{2} d u

= \int_{3}^{0} - (- u^{2} + 9) u^{2} d u

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

= - \int_{0}^{3} - (- u^{2} + 9) u^{2} d u

Take the constant out:

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

= - (- \int_{0}^{3} (- u^{2} + 9) u^{2} d u)

Expand

(- u^{2} + 9) u^{2} : - u^{4} + 9 u^{2}

= - (- \int_{0}^{3} - u^{4} + 9 u^{2} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{3} u^{4} d u + \int_{0}^{3} 9 u^{2} d u))

\int_{0}^{3} u^{4} d u = \frac{243}{5}

\int_{0}^{3} 9 u^{2} d u = 81

= - (- (- \frac{243}{5} + 81))

Simplify

- (- (- \frac{243}{5} + 81)) : \frac{162}{5}

= \frac{162}{5}

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What is the integral from 0 to 3 of x^3sqrt(9-x^2) ?
The integral from 0 to 3 of x^3sqrt(9-x^2) is 162/5