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

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

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

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

- 4

Solution steps

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

Expand

\frac{x - x}{x} : \frac{1}{x} - 1

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

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

\int_{1}^{9} 1 d x = 8

= 4 - 8

Simplify

= - 4

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