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

The integral from 0 to 4 of x^3sqrt(16-x^2) is 2048/15

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

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

\frac{2048}{15}

Decimal

136.53333 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{0} - (- u^{2} + 16) u^{2} d u

= \int_{4}^{0} - (- u^{2} + 16) u^{2} d u

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

= - \int_{0}^{4} - (- u^{2} + 16) u^{2} d u

Take the constant out:

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

= - (- \int_{0}^{4} (- u^{2} + 16) u^{2} d u)

Expand

(- u^{2} + 16) u^{2} : - u^{4} + 16 u^{2}

= - (- \int_{0}^{4} - u^{4} + 16 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}^{4} u^{4} d u + \int_{0}^{4} 16 u^{2} d u))

\int_{0}^{4} u^{4} d u = \frac{1024}{5}

\int_{0}^{4} 16 u^{2} d u = \frac{1024}{3}

= - (- (- \frac{1024}{5} + \frac{1024}{3}))

Simplify

- (- (- \frac{1024}{5} + \frac{1024}{3})) : \frac{2048}{15}

= \frac{2048}{15}

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What is the integral from 0 to 4 of x^3sqrt(16-x^2) ?
The integral from 0 to 4 of x^3sqrt(16-x^2) is 2048/15