What is the integral from 0 to 1-z-y of z ?

The integral from 0 to 1-z-y of z is ((-z+1-y)^2)/2

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integral from 0 to 1-z-y of z

\int_{0}^{1 - z - y} z d z

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

Solution steps

\int_{0}^{1 - z - y} z d z

Apply the Power Rule

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

= [\frac{z ^{2}}{2}]_{0}^{1 - y - z}

Compute the boundaries:

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

= \frac{1}{2} (- z + 1 - y)^{2}

Simplify

= \frac{( - z + 1 - y ) ^{2}}{2}

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What is the integral from 0 to 1-z-y of z ?
The integral from 0 to 1-z-y of z is ((-z+1-y)^2)/2