can i take the the 2 so it looks like this
= √2(1/√x)dx??
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
Hello, asilvester635!
$\displaystyle 18.\;\int^8_1\!\sqrt{\frac{2}{x}}\,dx$
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?
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