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}

Decimal

10.66666 \dots

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