What is the integral from 1 to 8 of sqrt(2/x) ?

The integral from 1 to 8 of sqrt(2/x) is 8-2sqrt(2)

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integral from 1 to 8 of sqrt(2/x)

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

8 - 22

Solution steps

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

Apply Integration By Parts

: [\frac{2}{x} x - \int - \frac{1}{2 x ^{\frac{1}{2}}} d x]_{1}^{8}

= [\frac{2}{x} x - \int - \frac{1}{2 x ^{\frac{1}{2}}} d x]_{1}^{8}

\int - \frac{1}{2 x ^{\frac{1}{2}}} d x = - 2 x

= [\frac{2}{x} x - (- 2 x)]_{1}^{8}

Simplify

= [\frac{2}{x} x + 2 x]_{1}^{8}

Compute the boundaries:

8 - 22

= 8 - 22

f (x) = - 6

x \to - 6 lim (\frac{1}{x + 6})

\frac{\partial}{\partial x} (4 x^{3} y^{2} + 4)

f (x) = 3 x^{4}

\frac{8}{s ^{3} ( s + 2 )}

What is the integral from 1 to 8 of sqrt(2/x) ?
The integral from 1 to 8 of sqrt(2/x) is 8-2sqrt(2)