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

The integral from 0 to 1-x of ((1-x-y)^2)/2 is ((-x+1)^3)/6

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integral from 0 to 1-x of ((1-x-y)^2)/2

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

\frac{( - x + 1 ) ^{3}}{6}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- x + 1}^{0} - u^{2} d u

= \frac{1}{2} \cdot \int_{- x + 1}^{0} - u^{2} d u

Take the constant out:

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

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

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- x + 1}^{0}

= \frac{1}{2} (- [\frac{u ^{3}}{3}]_{- x + 1}^{0})

Simplify

= - \frac{1}{2} [\frac{u ^{3}}{3}]_{- x + 1}^{0}

Compute the boundaries:

- \frac{( - x + 1 ) ^{3}}{3}

= - \frac{1}{2} (- \frac{( - x + 1 ) ^{3}}{3})

Simplify

= \frac{( - x + 1 ) ^{3}}{6}

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