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

The integral from 0 to 1 of (t-1/2)^2 is 1/12

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

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

\frac{1}{12}

Decimal

0.08333 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{- \frac{1}{2}}^{\frac{1}{2}}

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

Compute the boundaries:

\frac{1}{12}

= \frac{1}{12}

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What is the integral from 0 to 1 of (t-1/2)^2 ?
The integral from 0 to 1 of (t-1/2)^2 is 1/12