What is the integral from y^2 to 8-2y of 6y ?

The integral from y^2 to 8-2y of 6y is 3(-y^4+(-2y+8)^2)

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integral from y^2 to 8-2y of 6y

\int_{y^{2}}^{8 - 2 y} 6 y d y

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

Solution steps

\int_{y^{2}}^{8 - 2 y} 6 y d y

Take the constant out:

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

= 6 \cdot \int_{y^{2}}^{8 - 2 y} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{y^{2}}^{8 - 2 y}

= 6 [\frac{y ^{2}}{2}]_{y^{2}}^{8 - 2 y}

Compute the boundaries:

\frac{( 8 - 2 y ) ^{2}}{2} - \frac{y ^{4}}{2}

= 6 (\frac{( 8 - 2 y ) ^{2}}{2} - \frac{y ^{4}}{2})

Simplify

= 3 (- y^{4} + (- 2 y + 8)^{2})

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What is the integral from y^2 to 8-2y of 6y ?
The integral from y^2 to 8-2y of 6y is 3(-y^4+(-2y+8)^2)