What is the integral from 0 to 1 of ((1-x)^2)/4 ?

The integral from 0 to 1 of ((1-x)^2)/4 is 1/12

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integral from 0 to 1 of ((1-x)^2)/4

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{4} d x

\frac{1}{12}

Decimal Notation

0.08333 \dots

Solution steps

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{1} (1 - x)^{2} d x

Expand

(1 - x)^{2} : 1 - 2 x + x^{2}

= \frac{1}{4} \cdot \int_{0}^{1} 1 - 2 x + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int_{0}^{1} 1 d x - \int_{0}^{1} 2 x d x + \int_{0}^{1} x^{2} d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 2 x d x = 1

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{1}{4} (1 - 1 + \frac{1}{3})

Simplify

\frac{1}{4} (1 - 1 + \frac{1}{3}) : \frac{1}{12}

= \frac{1}{12}

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What is the integral from 0 to 1 of ((1-x)^2)/4 ?
The integral from 0 to 1 of ((1-x)^2)/4 is 1/12