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

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

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Expand

\frac{x - 4}{x} : x - \frac{4}{x}

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

\int_{1}^{9} x d x = \frac{52}{3}

\int_{1}^{9} \frac{4}{x} d x = 16

= \frac{52}{3} - 16

Simplify

\frac{52}{3} - 16 : \frac{4}{3}

= \frac{4}{3}

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