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

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

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

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

- \frac{40}{3}

Decimal

- 13.33333 \dots

Solution steps

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

Expand

\frac{x - 9}{x} : x - \frac{9}{x}

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

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

\int_{1}^{4} \frac{9}{x} d x = 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 (x-9)/(sqrt(x)) ?
The integral from 1 to 4 of (x-9)/(sqrt(x)) is -40/3