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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

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