What is the integral from 0 to 9-4y^2 of 16y ?

The integral from 0 to 9-4y^2 of 16y is 8(-4y^2+9)^2

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integral from 0 to 9-4y^2 of 16y

\int_{0}^{9 - 4 y^{2}} 16 y d y

8 (- 4 y^{2} + 9)^{2}

Solution steps

\int_{0}^{9 - 4 y^{2}} 16 y d y

Take the constant out:

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

= 16 \cdot \int_{0}^{9 - 4 y^{2}} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{9 - 4 y^{2}}

= 16 [\frac{y ^{2}}{2}]_{0}^{9 - 4 y^{2}}

Compute the boundaries:

\frac{1}{2} (- 4 y^{2} + 9)^{2}

= 16 \cdot \frac{1}{2} (- 4 y^{2} + 9)^{2}

Simplify

= 8 (- 4 y^{2} + 9)^{2}

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What is the integral from 0 to 9-4y^2 of 16y ?
The integral from 0 to 9-4y^2 of 16y is 8(-4y^2+9)^2