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

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

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integral from 0 to sqrt(4-y^2 of)z(-y+2)

\int_{0}^{4 - y^{2}} z (- y + 2) d z

- y^{2} + 4 - y \frac{- y ^{2} + 4}{2}

Solution steps

\int_{0}^{4 - y^{2}} z (- y + 2) d z

Take the constant out:

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

= (2 - y) \cdot \int_{0}^{4 - y^{2}} z d z

\int_{0}^{4 - y^{2}} z d z = \frac{- y ^{2} + 4}{2}

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

Simplify

(2 - y) \frac{- y ^{2} + 4}{2} : - y^{2} + 4 - y \frac{- y ^{2} + 4}{2}

= - y^{2} + 4 - y \frac{- y ^{2} + 4}{2}

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