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

The integral from 0 to 2 of (x^2-x+1) is 8/3

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

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

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

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

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

= \frac{8}{3} - 2 + 2

Simplify

= \frac{8}{3}

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What is the integral from 0 to 2 of (x^2-x+1) ?
The integral from 0 to 2 of (x^2-x+1) is 8/3