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

Decimal

5.17157 \dots

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) = x^{3}, at x = 3

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

\frac{\partial}{\partial x} (cos (x^{2}))

\int \frac{x ^{2} + 3 x - 2}{x} d x

\int \frac{2 x}{x ^{4} + 1} d x

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)