What is the integral from-6 to-3 of (u-3/(u^2)) ?

The integral from-6 to-3 of (u-3/(u^2)) is -14

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integral from-6 to-3 of (u-3/(u^2))

\int_{- 6}^{- 3} (u - \frac{3}{u ^{2}}) d u

- 14

Solution steps

\int_{- 6}^{- 3} u - \frac{3}{u ^{2}} 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_{- 6}^{- 3} u d u - \int_{- 6}^{- 3} \frac{3}{u ^{2}} d u

\int_{- 6}^{- 3} u d u = - \frac{27}{2}

\int_{- 6}^{- 3} \frac{3}{u ^{2}} d u = \frac{1}{2}

= - \frac{27}{2} - \frac{1}{2}

Simplify

- \frac{27}{2} - \frac{1}{2} : - 14

= - 14

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What is the integral from-6 to-3 of (u-3/(u^2)) ?
The integral from-6 to-3 of (u-3/(u^2)) is -14