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 13/12

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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{13}{12}

Decimal

1.08333 \dots

Solution steps

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

Simplify

(x + \frac{1}{2})^{2} : x^{2} + x + \frac{1}{4}

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

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

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

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

\int_{0}^{1} \frac{1}{4} d x = \frac{1}{4}

= \frac{1}{3} + \frac{1}{2} + \frac{1}{4}

Simplify

\frac{1}{3} + \frac{1}{2} + \frac{1}{4} : \frac{13}{12}

= \frac{13}{12}

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 13/12