What is the integral from 1 to 9 of (x-2)/(sqrt(x)) ?

The integral from 1 to 9 of (x-2)/(sqrt(x)) is 28/3

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integral from 1 to 9 of (x-2)/(sqrt(x))

\int_{1}^{9} \frac{x - 2}{x} d x

\frac{28}{3}

Decimal Notation

9.33333 \dots

Solution steps

\int_{1}^{9} \frac{x - 2}{x} d x

Expand

\frac{x - 2}{x} : x - \frac{2}{x}

= \int_{1}^{9} x - \frac{2}{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{2}{x} d x

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

\int_{1}^{9} \frac{2}{x} d x = 8

= \frac{52}{3} - 8

Simplify

\frac{52}{3} - 8 : \frac{28}{3}

= \frac{28}{3}

x \to 0 lim (\frac{x}{x ^{2} + 2 x})

\int_{0}^{2} x 7 - x^{2} d x

(11, 6), (16, y)

\frac{\partial}{\partial y} (y (e^{x} - 1))

\int (sec^{2} (x) - 1) d x

What is the integral from 1 to 9 of (x-2)/(sqrt(x)) ?
The integral from 1 to 9 of (x-2)/(sqrt(x)) is 28/3