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

The integral of (1-x-y)^2 is -((1-x-y)^3)/3+C

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integral of (1-x-y)^2

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

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

Solution steps

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

Apply u-substitution

: \int - u^{2} d u

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{u ^{3}}{3}

Substitute back

u = 1 - x - y

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

Add a constant to the solution

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

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