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

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

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

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

\frac{3 y ^{2} - 3 y + 1}{3}

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

= \frac{3 y ^{2} - 3 y + 1}{3}

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