What is the integral from 0 to 1 of (x^2-3x+7) ?

The integral from 0 to 1 of (x^2-3x+7) is 35/6

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integral from 0 to 1 of (x^2-3x+7)

\int_{0}^{1} (x^{2} - 3 x + 7) d x

\frac{35}{6}

Decimal

5.83333 \dots

Solution steps

\int_{0}^{1} x^{2} - 3 x + 7 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} 3 x d x + \int_{0}^{1} 7 d x

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

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

\int_{0}^{1} 7 d x = 7

= \frac{1}{3} - \frac{3}{2} + 7

Simplify

\frac{1}{3} - \frac{3}{2} + 7 : \frac{35}{6}

= \frac{35}{6}

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What is the integral from 0 to 1 of (x^2-3x+7) ?
The integral from 0 to 1 of (x^2-3x+7) is 35/6