What is the integral from 0 to 4-z^2 of z ?

The integral from 0 to 4-z^2 of z is ((-z^2+4)^2)/2

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integral from 0 to 4-z^2 of z

\int_{0}^{4 - z^{2}} z d z

\frac{( - z ^{2} + 4 ) ^{2}}{2}

Solution steps

\int_{0}^{4 - z^{2}} z d z

Apply the Power Rule

: [\frac{z ^{2}}{2}]_{0}^{4 - z^{2}}

= [\frac{z ^{2}}{2}]_{0}^{4 - z^{2}}

Compute the boundaries:

\frac{1}{2} (- z^{2} + 4)^{2}

= \frac{1}{2} (- z^{2} + 4)^{2}

Simplify

= \frac{( - z ^{2} + 4 ) ^{2}}{2}

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What is the integral from 0 to 4-z^2 of z ?
The integral from 0 to 4-z^2 of z is ((-z^2+4)^2)/2