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

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

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

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

\frac{1}{12}

Decimal

0.08333 \dots

Solution steps

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

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}

\frac{d y}{d t} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} y = (3 x + 2) (5 x^{2} + 4)

\int \frac{x}{x - 1 + 1} d x

f (x) = 2 e^{- x} x - e^{- x} x^{2}

\frac{x ^{2}}{x + 5}

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