# Thread: evaluate the definite integral

1. ## evaluate the definite integral

can i take the the 2 so it looks like this

= √2(1/√x)dx??

2. ## Re: evaluate the definite integral

Originally Posted by asilvester635
can i take the the 2 so it looks like this
= √2(1/√x)dx??

What is the derivative of $\sqrt{2x}~?$

3. ## Re: evaluate the definite integral

it will be x^-1/2

4. ## Re: evaluate the definite integral

Originally Posted by asilvester635
it will be x^-1/2
No it is not. If you cannot do derivatives, the integrals are impossible for you.

5. ## Re: evaluate the definite integral

do i rewrite this?

6. ## Re: evaluate the definite integral

Originally Posted by asilvester635
do i rewrite this?
Look at this webpage.

7. ## Re: evaluate the definite integral

The answer to the original question is yes, $\int_1^8\sqrt\frac{2}{x}\ dx = \sqrt{2}\int_1^8\sqrt\frac{1}{x}\ dx$.

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

- Hollywood

8. ## Re: evaluate the definite integral

Hello, asilvester635!

$18.\;\int^8_1\!\sqrt{\frac{2}{x}}\,dx$
Can i take the the 2 so it looks like this? . $\sqrt{2}\left(\frac{1}{\sqrt{x}}\right)$ . Yes!

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

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

Got it?

9. ## Re: evaluate the definite integral

Originally Posted by asilvester635
can i take the the 2 so it looks like this

= √2(1/√x)dx??
You can find the answer to this definite integral like this:

\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*}

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

You can check the answer at:

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