What is the integral from 0 to 5-2x-1/2 y of 6z^2 ?

The integral from 0 to 5-2x-1/2 y of 6z^2 is ((10-4x-y)^3)/4

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integral from 0 to 5-2x-1/2 y of 6z^2

\int_{0}^{5 - 2 x - \frac{1}{2} y} 6 z^{2} d z

\frac{( 10 - 4 x - y ) ^{3}}{4}

Solution steps

\int_{0}^{5 - 2 x - \frac{1}{2} y} 6 z^{2} d z

Take the constant out:

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

= 6 \cdot \int_{0}^{5 - 2 x - \frac{1}{2} y} z^{2} d z

Apply the Power Rule

: [\frac{z ^{3}}{3}]_{0}^{5 - 2 x - \frac{1}{2} y}

= 6 [\frac{z ^{3}}{3}]_{0}^{5 - 2 x - \frac{1}{2} y}

Compute the boundaries:

\frac{( 10 - 4 x - y ) ^{3}}{24}

= 6 \cdot \frac{( 10 - 4 x - y ) ^{3}}{24}

Simplify

= \frac{( 10 - 4 x - y ) ^{3}}{4}

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What is the integral from 0 to 5-2x-1/2 y of 6z^2 ?
The integral from 0 to 5-2x-1/2 y of 6z^2 is ((10-4x-y)^3)/4