What is the integral from 0 to 1-x of 6(1-x-y) ?

The integral from 0 to 1-x of 6(1-x-y) is 6(x^2-2x-((-x+1)^2)/2+1)

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integral from 0 to 1-x of 6(1-x-y)

\int_{0}^{1 - x} 6 (1 - x - y) d y

6 (x^{2} - 2 x - \frac{( - x + 1 ) ^{2}}{2} + 1)

Solution steps

\int_{0}^{1 - x} 6 (1 - x - y) d y

Take the constant out:

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

= 6 \cdot \int_{0}^{1 - x} 1 - 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

= 6 (\int_{0}^{1 - x} 1 d y - \int_{0}^{1 - x} x d y - \int_{0}^{1 - x} y d y)

\int_{0}^{1 - x} 1 d y = - x + 1

\int_{0}^{1 - x} x d y = x (- x + 1)

\int_{0}^{1 - x} y d y = \frac{( - x + 1 ) ^{2}}{2}

= 6 (- x + 1 - x (- x + 1) - \frac{( - x + 1 ) ^{2}}{2})

Expand

- x + 1 - x (- x + 1) - \frac{( - x + 1 ) ^{2}}{2} : x^{2} - 2 x - \frac{( - x + 1 ) ^{2}}{2} + 1

= 6 (x^{2} - 2 x - \frac{( - x + 1 ) ^{2}}{2} + 1)

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What is the integral from 0 to 1-x of 6(1-x-y) ?
The integral from 0 to 1-x of 6(1-x-y) is 6(x^2-2x-((-x+1)^2)/2+1)