What is the integral from 0 to 2 of (x-4)(x-1) ?

The integral from 0 to 2 of (x-4)(x-1) is 2/3

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integral from 0 to 2 of (x-4)(x-1)

\int_{0}^{2} (x - 4) (x - 1) d x

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

\int_{0}^{2} (x - 4) (x - 1) d x

Expand

(x - 4) (x - 1) : x^{2} - 5 x + 4

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 5 x d x = 10

\int_{0}^{2} 4 d x = 8

= \frac{8}{3} - 10 + 8

Simplify

\frac{8}{3} - 10 + 8 : \frac{2}{3}

= \frac{2}{3}

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What is the integral from 0 to 2 of (x-4)(x-1) ?
The integral from 0 to 2 of (x-4)(x-1) is 2/3