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

The integral from 0 to 1 of (x^2-3x+8) is 41/6

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

\int_{0}^{1} (x^{2} - 3 x + 8) d x

\frac{41}{6}

Decimal

6.83333 \dots

Solution steps

\int_{0}^{1} x^{2} - 3 x + 8 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} 8 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} 8 d x = 8

= \frac{1}{3} - \frac{3}{2} + 8

Simplify

\frac{1}{3} - \frac{3}{2} + 8 : \frac{41}{6}

= \frac{41}{6}

What is the integral from 0 to 1 of (x^2-3x+8) ?
The integral from 0 to 1 of (x^2-3x+8) is 41/6