What is the integral from 0 to 2 of rsqrt(16-r^2) ?

The integral from 0 to 2 of rsqrt(16-r^2) is (8(8-3sqrt(3)))/3

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integral from 0 to 2 of rsqrt(16-r^2)

\int_{0}^{2} r 16 - r^{2} d r

\frac{8 ( 8 - 3 3 )}{3}

Decimal

7.47692 \dots

Solution steps

\int_{0}^{2} r 16 - r^{2} d r

Apply u-substitution

: \int_{16}^{12} - \frac{u}{2} d u

= \int_{16}^{12} - \frac{u}{2} d u

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

= - \int_{12}^{16} - \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_{12}^{16} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16}

Compute the boundaries:

\frac{128}{3} - 163

= \frac{1}{2} (\frac{128}{3} - 163)

Simplify

= \frac{8 ( 8 - 3 3 )}{3}

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\int_{- 1}^{1} (8 x^{3} - 6 x + 1) d x

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What is the integral from 0 to 2 of rsqrt(16-r^2) ?
The integral from 0 to 2 of rsqrt(16-r^2) is (8(8-3sqrt(3)))/3