What is the integral from-1 to 3 of (3u-5)(u+2) ?

The integral from-1 to 3 of (3u-5)(u+2) is -8

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integral from-1 to 3 of (3u-5)(u+2)

\int_{- 1}^{3} (3 u - 5) (u + 2) d u

- 8

Solution steps

\int_{- 1}^{3} (3 u - 5) (u + 2) d u

Expand

(3 u - 5) (u + 2) : 3 u^{2} + u - 10

= \int_{- 1}^{3} 3 u^{2} + u - 10 d u

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}^{3} 3 u^{2} d u + \int_{- 1}^{3} u d u - \int_{- 1}^{3} 10 d u

\int_{- 1}^{3} 3 u^{2} d u = 28

\int_{- 1}^{3} u d u = 4

\int_{- 1}^{3} 10 d u = 40

= 28 + 4 - 40

Simplify

= - 8

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What is the integral from-1 to 3 of (3u-5)(u+2) ?
The integral from-1 to 3 of (3u-5)(u+2) is -8