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

The integral from 0 to 1 of (x^2-x+1) is 5/6

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

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

\frac{5}{6}

Decimal

0.83333 \dots

Solution steps

\int_{0}^{1} x^{2} - x + 1 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} 1 d x

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

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

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

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

Simplify

\frac{1}{3} - \frac{1}{2} + 1 : \frac{5}{6}

= \frac{5}{6}

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What is the integral from 0 to 1 of (x^2-x+1) ?
The integral from 0 to 1 of (x^2-x+1) is 5/6