# Thread: Evaluate the definite integral of the algebraic function

1. ## 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

2. ## 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)

3. ## 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)????

4. ## Re: Evaluate the definite integral of the algebraic function

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.

$n \ne - 1,\quad \int {x^n dx} = \frac{{x^{n + 1} }}{{n + 1}}$

5. ## 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

6. ## 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????

7. ## Re: Evaluate the definite integral of the algebraic function

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

8. ## 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?

9. ## Re: Evaluate the definite integral of the algebraic function

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 $2\sqrt{2}\sqrt{x}+C$

Or we could write it as $\sqrt{8x}+C$

10. ## 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?

11. ## Re: Evaluate the definite integral of the algebraic function

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?

$2\sqrt{2}=\sqrt{4}\sqrt{2}=\sqrt{8}$

12. ## Re: Evaluate the definite integral of the algebraic function

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: $2 \sqrt{2} = 2^1 \cdot 2^{1/2} = 2^{3/2} = \left ( 2^3 \right )^{1/2} = \left ( 2^3 \right ) ^{1/2} = \sqrt{8}$

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.

13. ## Re: Evaluate the definite integral of the algebraic function

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

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

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

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

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

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

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

$8 - 2\sqrt{2}$