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

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

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

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

- \frac{8}{3}

Decimal Notation

- 2.66666 \dots

Solution steps

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

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

\int_{0}^{1} 8 x d x = 4

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

= \frac{1}{3} - 4 + 1

Simplify

\frac{1}{3} - 4 + 1 : - \frac{8}{3}

= - \frac{8}{3}

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