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

The integral from 1 to 5 of (x^2-6x+5) is -32/3

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integral from 1 to 5 of (x^2-6x+5)

\int_{1}^{5} (x^{2} - 6 x + 5) d x

- \frac{32}{3}

Solution steps

\int_{1}^{5} x^{2} - 6 x + 5 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_{1}^{5} x^{2} d x - \int_{1}^{5} 6 x d x + \int_{1}^{5} 5 d x

\int_{1}^{5} x^{2} d x = \frac{124}{3}

\int_{1}^{5} 6 x d x = 72

\int_{1}^{5} 5 d x = 20

= \frac{124}{3} - 72 + 20

Simplify

\frac{124}{3} - 72 + 20 : - \frac{32}{3}

= - \frac{32}{3}

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What is the integral from 1 to 5 of (x^2-6x+5) ?
The integral from 1 to 5 of (x^2-6x+5) is -32/3