# Evaluating an Integral

• Feb 25th 2011, 11:17 AM
kikiya
Evaluating an Integral

e^(x)^1/2/ (x^1/2)

I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please give me some insight on this! (Wondering)
• Feb 25th 2011, 11:20 AM
TheEmptySet
Quote:

Originally Posted by kikiya

e^(x)^1/2/ (x^1/2)

I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please five me some insight on this! (Wondering)

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx$

$\displaystyle u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}$

If you sub this in you should get

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du$
• Feb 25th 2011, 11:21 AM
pickslides
You're substitution is correct, what did you find $\displaystyle \displaystyle \frac{du}{dx}$ to be?

Edit: TES beat me to the punch.
• Feb 25th 2011, 11:24 AM
pickslides
Quote:

Originally Posted by TheEmptySet

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx$

$\displaystyle u=\sqrt{x} \implies du=\frac{dx}{\sqrt{x}}$

If you sub this in you should get

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int e^{u}du$

I get

$\displaystyle \displaystyle u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}$

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du$
• Feb 25th 2011, 11:30 AM
TheEmptySet
Quote:

Originally Posted by pickslides
I get

$\displaystyle \displaystyle u=\sqrt{x} \implies du=\frac{dx}{2\sqrt{x}}$

$\displaystyle \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=2\int e^{u}du$

and you are correct!
• Feb 25th 2011, 11:38 AM
kikiya
I'm sorry, but where did you get the 2 that's outside of the integral? Please explain your process lol
• Feb 25th 2011, 11:58 AM
pickslides
O.K given $\displaystyle \displaystyle u = \sqrt{x}$what do you get for $\displaystyle \displaystyle \frac{du}{dx}$ ?
• Feb 25th 2011, 01:48 PM
kikiya
Sorry I'm late but is it 1/ 2 (x^1/2) ?
• Feb 25th 2011, 02:25 PM
NOX Andrew
Exactly. Unfortunately, there isn't a 1/2 inside the integral, so you can't substitute $\displaystyle du = \frac{1}{2\sqrt{x}}dx$ yet. Fortunately, you can place a 1/2 inside the integral as long as you place a 2 outside the integral (to balance the 1/2). Now, the problem looks like:

$\displaystyle 2\int \frac{e^{\sqrt{x}}}{2\sqrt{x}} \, dx$

Substituting $\displaystyle u = \sqrt{x}$ and $\displaystyle du = \frac{1}{2\sqrt{x}}dx$ gives:

$\displaystyle 2\int e^u \, du$
• Feb 26th 2011, 03:06 AM
HallsofIvy
Quote:

Originally Posted by kikiya

e^(x)^1/2/ (x^1/2)

Strictly speaking, what you wrote here was
$\displaystyle \frac{\left(e^x)^{1/2}}{x^{1/2}}= \frac{e^{\frac{x}{2}}}{x^{1/2}}$
but apparently you meant
$\displaystyle e^{x^{1/2}}}{x^{1/2}}$
which would be e^(x^(1/2))/x^(1/2)

Quote:

I started out with u substitution. First I let radical x be u, but I couldn't work it out properly because I didn't know what to cross out. Please give me some insight on this! (Wondering)
• Feb 26th 2011, 08:05 AM
kikiya
Oh, I see the error! Thank you! :)
• Feb 26th 2011, 08:07 AM
kikiya
Thank you for walking me through it! I was confiused about the balancing out, but now I get it! Thank you! :D