What is the integral from 0 to 4 of (x-5)(x-1) ?

The integral from 0 to 4 of (x-5)(x-1) is -20/3

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integral from 0 to 4 of (x-5)(x-1)

\int_{0}^{4} (x - 5) (x - 1) d x

- \frac{20}{3}

Solution steps

\int_{0}^{4} (x - 5) (x - 1) d x

Expand

(x - 5) (x - 1) : x^{2} - 6 x + 5

= \int_{0}^{4} 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_{0}^{4} x^{2} d x - \int_{0}^{4} 6 x d x + \int_{0}^{4} 5 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 6 x d x = 48

\int_{0}^{4} 5 d x = 20

= \frac{64}{3} - 48 + 20

Simplify

\frac{64}{3} - 48 + 20 : - \frac{20}{3}

= - \frac{20}{3}

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What is the integral from 0 to 4 of (x-5)(x-1) ?
The integral from 0 to 4 of (x-5)(x-1) is -20/3