What is the integral from y-2 to 4-y^2 of x ?

The integral from y-2 to 4-y^2 of x is (y^4-9y^2+4y+12)/2

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integral from y-2 to 4-y^2 of x

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

\frac{y ^{4} - 9 y ^{2} + 4 y + 12}{2}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

= \frac{y ^{4} - 9 y ^{2} + 4 y + 12}{2}

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What is the integral from y-2 to 4-y^2 of x ?
The integral from y-2 to 4-y^2 of x is (y^4-9y^2+4y+12)/2