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

The integral from 4-y to 4-y^2 of y is (y^4-9y^2+8y)/2

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

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

\frac{y ^{4} - 9 y ^{2} + 8 y}{2}

Solution steps

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

= \frac{y ^{4} - 9 y ^{2} + 8 y}{2}

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What is the integral from 4-y to 4-y^2 of y ?
The integral from 4-y to 4-y^2 of y is (y^4-9y^2+8y)/2