What is the integral from 0 to 1 of (x^2-x+1/6)^2 ?

The integral from 0 to 1 of (x^2-x+1/6)^2 is 1/180

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integral from 0 to 1 of (x^2-x+1/6)^2

\int_{0}^{1} (x^{2} - x + \frac{1}{6})^{2} d x

\frac{1}{180}

Decimal

0.00555 \dots

Solution steps

\int_{0}^{1} (x^{2} - x + \frac{1}{6})^{2} d x

Expand

(x^{2} - x + \frac{1}{6})^{2} : \frac{1}{36} + x^{4} - 2 x^{3} + \frac{4 x ^{2}}{3} - \frac{x}{3}

= \int_{0}^{1} \frac{1}{36} + x^{4} - 2 x^{3} + \frac{4 x ^{2}}{3} - \frac{x}{3} 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} \frac{1}{36} d x + \int_{0}^{1} x^{4} d x - \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} \frac{4 x ^{2}}{3} d x - \int_{0}^{1} \frac{x}{3} d x

\int_{0}^{1} \frac{1}{36} d x = \frac{1}{36}

\int_{0}^{1} x^{4} d x = \frac{1}{5}

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

\int_{0}^{1} \frac{4 x ^{2}}{3} d x = \frac{4}{9}

\int_{0}^{1} \frac{x}{3} d x = \frac{1}{6}

= \frac{1}{36} + \frac{1}{5} - \frac{1}{2} + \frac{4}{9} - \frac{1}{6}

Simplify

\frac{1}{36} + \frac{1}{5} - \frac{1}{2} + \frac{4}{9} - \frac{1}{6} : \frac{1}{180}

= \frac{1}{180}

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What is the integral from 0 to 1 of (x^2-x+1/6)^2 ?
The integral from 0 to 1 of (x^2-x+1/6)^2 is 1/180