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

The integral from 0 to 2 of r^3sqrt(4-r^2) is 64/15

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

\int_{0}^{2} r^{3} 4 - r^{2} d r

\frac{64}{15}

Decimal

4.26666 \dots

Solution steps

\int_{0}^{2} r^{3} 4 - r^{2} d r

Apply u-substitution

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

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

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

Take the constant out:

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

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

Expand

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

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

\int_{0}^{2} u^{4} d u = \frac{32}{5}

\int_{0}^{2} 4 u^{2} d u = \frac{32}{3}

= - (- (- \frac{32}{5} + \frac{32}{3}))

Simplify

- (- (- \frac{32}{5} + \frac{32}{3})) : \frac{64}{15}

= \frac{64}{15}

What is the integral from 0 to 2 of r^3sqrt(4-r^2) ?
The integral from 0 to 2 of r^3sqrt(4-r^2) is 64/15