What is the integral from 0 to 2 of rsqrt(9-r^2) ?

The integral from 0 to 2 of rsqrt(9-r^2) is (27-5sqrt(5))/3

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integral from 0 to 2 of rsqrt(9-r^2)

\int_{0}^{2} r 9 - r^{2} d r

\frac{27 - 5 5}{3}

Decimal

5.27322 \dots

Solution steps

\int_{0}^{2} r 9 - r^{2} d r

Apply u-substitution

: \int_{9}^{5} - \frac{u}{2} d u

= \int_{9}^{5} - \frac{u}{2} d u

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

= - \int_{5}^{9} - \frac{u}{2} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{5}^{9} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{9}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{9})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{9}

Compute the boundaries:

18 - \frac{10 5}{3}

= \frac{1}{2} (18 - \frac{10 5}{3})

Simplify

= \frac{27 - 5 5}{3}

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What is the integral from 0 to 2 of rsqrt(9-r^2) ?
The integral from 0 to 2 of rsqrt(9-r^2) is (27-5sqrt(5))/3