can i take the the 2 so it looks like this

= √2(1/√x)dx??

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- Dec 14th 2012, 02:45 PMasilvester635evaluate the definite integral
can i take the the 2 so it looks like this

= √2(1/√x)dx?? - Dec 14th 2012, 02:59 PMPlatoRe: evaluate the definite integral
- Dec 14th 2012, 03:00 PMasilvester635Re: evaluate the definite integral
it will be x^-1/2

- Dec 14th 2012, 03:03 PMPlatoRe: evaluate the definite integral
- Dec 14th 2012, 03:09 PMasilvester635Re: evaluate the definite integral
do i rewrite this?

- Dec 14th 2012, 03:16 PMPlatoRe: evaluate the definite integral
- Dec 14th 2012, 05:33 PMhollywoodRe: evaluate the definite integral
The answer to the original question is yes, $\displaystyle \int_1^8\sqrt\frac{2}{x}\ dx = \sqrt{2}\int_1^8\sqrt\frac{1}{x}\ dx$.

Plato has a point - you need some proficiency with derivatives before you try to do integrals.

- Hollywood - Dec 14th 2012, 06:09 PMSorobanRe: evaluate the definite integral
Hello, asilvester635!

Quote:

$\displaystyle 18.\;\int^8_1\!\sqrt{\frac{2}{x}}\,dx$

Can i take the the 2 so it looks like this? .$\displaystyle \sqrt{2}\left(\frac{1}{\sqrt{x}}\right)$ . Yes!

$\displaystyle \sqrt{\frac{2}{x}} \;=\;\frac{\sqrt{2}}{\sqrt{x}} \;=\;\sqrt{2}\cdot\frac{1}{\sqrt{x}} \;=\;2x^{-\frac{1}{2}} $

Therefore: .$\displaystyle \int^8_1\!\sqrt{\frac{2}{x}}\,dx \;=\;\sqrt{2}\!\int^8_1 x^{-\frac{1}{2}}\,dx$

Got it?

- Dec 17th 2012, 02:38 PMx3bnmRe: evaluate the definite integral
You can find the answer to this definite integral like this:

$\displaystyle \begin{align*}\int^8_1\!\sqrt{\frac{2}{x}}\,dx =& 2\sqrt{2}\sqrt{x}\Big{]}^{8}_{1} \\ =& 2\sqrt{2} \sqrt{8} - 2\sqrt{2}\sqrt{1} \\ =& 8 - 2\sqrt{2}\end{align*}$

$\displaystyle \therefore \int^8_1\!\sqrt{\frac{2}{x}}\,dx = 8 - 2\sqrt{2}$

You can check the answer at:

int (2)^(1/2) 1/(x)^(1/2) dx x=1 to 8 - Wolfram|Alpha