What is the integral from 1 to 4 of (u-9)/(sqrt(u)) ?

The integral from 1 to 4 of (u-9)/(sqrt(u)) is -40/3

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integral from 1 to 4 of (u-9)/(sqrt(u))

\int_{1}^{4} \frac{u - 9}{u} d u

- \frac{40}{3}

Decimal Notation

- 13.33333 \dots

Solution steps

\int_{1}^{4} \frac{u - 9}{u} d u

Expand

\frac{u - 9}{u} : u - \frac{9}{u}

= \int_{1}^{4} u - \frac{9}{u} 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}^{4} u d u - \int_{1}^{4} \frac{9}{u} d u

\int_{1}^{4} u d u = \frac{14}{3}

\int_{1}^{4} \frac{9}{u} d u = 18

= \frac{14}{3} - 18

Simplify

\frac{14}{3} - 18 : - \frac{40}{3}

= - \frac{40}{3}

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What is the integral from 1 to 4 of (u-9)/(sqrt(u)) ?
The integral from 1 to 4 of (u-9)/(sqrt(u)) is -40/3