What is the integral from 0 to 3 of rsqrt(25-r^2) ?

The integral from 0 to 3 of rsqrt(25-r^2) is 61/3

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integral from 0 to 3 of rsqrt(25-r^2)

\int_{0}^{3} r 25 - r^{2} d r

\frac{61}{3}

Decimal Notation

20.33333 \dots

Solution steps

\int_{0}^{3} r 25 - r^{2} d r

Apply u-substitution

: \int_{25}^{16} - \frac{u}{2} d u

= \int_{25}^{16} - \frac{u}{2} d u

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

= - \int_{16}^{25} - \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_{16}^{25} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{25}

Compute the boundaries:

\frac{122}{3}

= \frac{1}{2} \cdot \frac{122}{3}

Simplify

= \frac{61}{3}

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What is the integral from 0 to 3 of rsqrt(25-r^2) ?
The integral from 0 to 3 of rsqrt(25-r^2) is 61/3