What is the integral from 0 to 6-x of (6-x-z) ?

The integral from 0 to 6-x of (6-x-z) is x^2-12x-((-x+6)^2)/2+36

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integral from 0 to 6-x of (6-x-z)

\int_{0}^{6 - x} (6 - x - z) d z

x^{2} - 12 x - \frac{( - x + 6 ) ^{2}}{2} + 36

Solution steps

\int_{0}^{6 - x} 6 - x - z d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{6 - x} 6 d z - \int_{0}^{6 - x} x d z - \int_{0}^{6 - x} z d z

\int_{0}^{6 - x} 6 d z = 6 (- x + 6)

\int_{0}^{6 - x} x d z = x (- x + 6)

\int_{0}^{6 - x} z d z = \frac{( - x + 6 ) ^{2}}{2}

= 6 (- x + 6) - x (- x + 6) - \frac{( - x + 6 ) ^{2}}{2}

Simplify

6 (- x + 6) - x (- x + 6) - \frac{( - x + 6 ) ^{2}}{2} : x^{2} - 12 x - \frac{( - x + 6 ) ^{2}}{2} + 36

= x^{2} - 12 x - \frac{( - x + 6 ) ^{2}}{2} + 36

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What is the integral from 0 to 6-x of (6-x-z) ?
The integral from 0 to 6-x of (6-x-z) is x^2-12x-((-x+6)^2)/2+36