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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int_{- a}^{1 - a} u^{2} d u

= \int_{- a}^{1 - a} u^{2} d u

Apply the Power Rule

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

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

Compute the boundaries:

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

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

Simplify

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

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