What is the integral from 0 to 4-x of (4-x-y) ?

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

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integral from 0 to 4-x of (4-x-y)

\int_{0}^{4 - x} (4 - x - y) d y

x^{2} - 8 x - \frac{( - x + 4 ) ^{2}}{2} + 16

Solution steps

\int_{0}^{4 - x} 4 - x - y d y

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}^{4 - x} 4 d y - \int_{0}^{4 - x} x d y - \int_{0}^{4 - x} y d y

\int_{0}^{4 - x} 4 d y = 4 (- x + 4)

\int_{0}^{4 - x} x d y = x (- x + 4)

\int_{0}^{4 - x} y d y = \frac{( - x + 4 ) ^{2}}{2}

= 4 (- x + 4) - x (- x + 4) - \frac{( - x + 4 ) ^{2}}{2}

Simplify

4 (- x + 4) - x (- x + 4) - \frac{( - x + 4 ) ^{2}}{2} : x^{2} - 8 x - \frac{( - x + 4 ) ^{2}}{2} + 16

= x^{2} - 8 x - \frac{( - x + 4 ) ^{2}}{2} + 16

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