What is the integral from 0 to 5 of zsqrt(25-z^2) ?

The integral from 0 to 5 of zsqrt(25-z^2) is 125/3

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integral from 0 to 5 of zsqrt(25-z^2)

\int_{0}^{5} z 25 - z^{2} d z

\frac{125}{3}

Solution steps

\int_{0}^{5} z 25 - z^{2} d z

Apply u-substitution

: \int_{25}^{0} - \frac{u}{2} d u

= \int_{25}^{0} - \frac{u}{2} d u

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

= - \int_{0}^{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_{0}^{25} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25})

Simplify

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{25}

Compute the boundaries:

\frac{250}{3}

= \frac{1}{2} \cdot \frac{250}{3}

Simplify

= \frac{125}{3}

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What is the integral from 0 to 5 of zsqrt(25-z^2) ?
The integral from 0 to 5 of zsqrt(25-z^2) is 125/3