# Evaluate the definite integral of the algebraic function

• Dec 3rd 2012, 06:51 AM
asilvester635
Evaluate the definite integral of the algebraic function
If you integrate the square root of 2/x does it come out to just 2x because when i rewrite it its going to be the square root of 2x^-1
Then i turn the square root into an exponent which looks like this 2x^-1+2.... -1 + 2 = 1.... So i have just 1 as an exponent.... now i just have 2x
• Dec 3rd 2012, 07:16 AM
coolge
Re: Evaluate the definite integral of the algebraic function
integral of (1/sqrt x) = integral (x ^(-1/2)} = x^(-1/2 + 1)/(-1/2 + 1) = sqrt(x)/(1/2)
• Dec 3rd 2012, 07:27 AM
asilvester635
Re: Evaluate the definite integral of the algebraic function
ok i understand how you got x^-1/2 part but how did you get (-1/2 + 1)/(-1/2 + 1)????
• Dec 3rd 2012, 07:40 AM
Plato
Re: Evaluate the definite integral of the algebraic function
Quote:

Originally Posted by asilvester635
ok i understand how you got x^-1/2 part but how did you get (-1/2 + 1)/(-1/2 + 1)????

You ought to know the basics.

$\displaystyle n \ne - 1,\quad \int {x^n dx} = \frac{{x^{n + 1} }}{{n + 1}}$
• Dec 3rd 2012, 07:41 AM
coolge
Re: Evaluate the definite integral of the algebraic function
integral of x^n ={1/(n+1)}*{ x^{n+1)}. Put n = -1/2 you get (1/2)*sqrt x
• Dec 3rd 2012, 09:29 AM
asilvester635
Re: Evaluate the definite integral of the algebraic function
i know the basics...... im just asking how you would rewrite √2/x??? isn't it √2x^-1????
• Dec 3rd 2012, 10:15 AM
Plato
Re: Evaluate the definite integral of the algebraic function
Quote:

Originally Posted by asilvester635
i know the basics...... im just asking how you would rewrite √2/x??? isn't it √2x^-1????

"i know the basics".Do you really?

Basic algebra: $\displaystyle \sqrt{\frac{2}{x}}=\sqrt{2}x^{-\frac{1}{2}}$
• Dec 3rd 2012, 11:13 AM
asilvester635
Re: Evaluate the definite integral of the algebraic function
yes that what i got but i put -1 instead of putting -1/2.... NOW WHAT DO I DO with the square root?
• Dec 3rd 2012, 11:35 AM
Plato
Re: Evaluate the definite integral of the algebraic function
Quote:

Originally Posted by asilvester635
yes that what i got but i put -1 instead of putting -1/2.... NOW WHAT DO I DO with the square root?

The final answer to the integral is $\displaystyle 2\sqrt{2}\sqrt{x}+C$

Or we could write it as $\displaystyle \sqrt{8x}+C$
• Dec 3rd 2012, 04:14 PM
asilvester635
Re: Evaluate the definite integral of the algebraic function
oh it makes sense now but how is 2√2√x equal to √8x?
And since this is definite integral we can remove the + C right?
• Dec 3rd 2012, 04:24 PM
Plato
Re: Evaluate the definite integral of the algebraic function
Quote:

Originally Posted by asilvester635
oh it makes sense now but how is 2√2√x equal to √8x?
And since this is definite integral we can remove the + C right?

$\displaystyle 2\sqrt{2}=\sqrt{4}\sqrt{2}=\sqrt{8}$
• Dec 3rd 2012, 04:28 PM
topsquark
Re: Evaluate the definite integral of the algebraic function
Quote:

Originally Posted by asilvester635
oh it makes sense now but how is 2√2√x equal to √8x?
And since this is definite integral we can remove the + C right?

Your problem seems to me like you need to do the indefinite integral. You should be able to put the limits in yourself. As for the other question: $\displaystyle 2 \sqrt{2} = 2^1 \cdot 2^{1/2} = 2^{3/2} = \left ( 2^3 \right )^{1/2} = \left ( 2^3 \right ) ^{1/2} = \sqrt{8}$

You might need a review of multiplication of radicals, which is a pre-Calculus topic.

-Dan

@Plato: Sorry for the snipe. I thought you were gone. (Nod)
• Dec 3rd 2012, 04:55 PM
skeeter
Re: Evaluate the definite integral of the algebraic function
$\displaystyle \int_1^8 \sqrt{\frac{2}{x}} \, dx$

$\displaystyle \sqrt{2} \int_1^8 \frac{1}{\sqrt{x}} \, dx$

$\displaystyle 2\sqrt{2} \int_1^8 \frac{1}{2\sqrt{x}} \, dx$

$\displaystyle \frac{1}{2\sqrt{x}}$ is the derivative of $\displaystyle \sqrt{x}$ ...

$\displaystyle 2\sqrt{2} \left[\sqrt{x}\right]_1^8$

$\displaystyle 2\sqrt{2} \left[\sqrt{8} - \sqrt{1} \right]$

$\displaystyle 2\sqrt{16} - 2\sqrt{2}$

$\displaystyle 8 - 2\sqrt{2}$