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}

\frac{\partial}{\partial x} (x + 5 y)

\frac{x}{3 + x}

arcsec (2 x + 3)

\frac{d}{d x} (ln (2 x^{3} + 8 x))

\int 8^{2 x + 6} d x

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