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

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

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

\int_{0}^{4 - y^{2}} 2 x y d x

y (- y^{2} + 4)

Solution steps

\int_{0}^{4 - y^{2}} 2 x y d x

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}} x d x

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

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