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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

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