Evaluating an Integral

**kikiya** · February 25th 2011, 11:17 AM

Please help me evaluate

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!

**TheEmptySet** · February 25th 2011, 11:20 AM

Originally Posted by kikiya

Please help me evaluate

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!

Your substitution is correct.

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

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

If you sub this in you should get

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

**pickslides** · February 25th 2011, 11:21 AM

You're substitution is correct, what did you find $\displaystyle \frac{du}{dx}$ to be?

Edit: TES beat me to the punch.

**pickslides** · February 25th 2011, 11:24 AM

Originally Posted by TheEmptySet

Your substitution is correct.

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

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

If you sub this in you should get

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

I get

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

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

**TheEmptySet** · February 25th 2011, 11:30 AM

Originally Posted by pickslides

I get

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

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

and you are correct!

**kikiya** · February 25th 2011, 11:38 AM

I'm sorry, but where did you get the 2 that's outside of the integral? Please explain your process lol

**pickslides** · February 25th 2011, 11:58 AM

O.K given $\displaystyle u = \sqrt{x}$ what do you get for $\displaystyle \frac{du}{dx}$ ?

**kikiya** · February 25th 2011, 01:48 PM

Sorry I'm late but is it 1/ 2 (x^1/2) ?

**NOX Andrew** · February 25th 2011, 02:25 PM

Exactly. Unfortunately, there isn't a 1/2 inside the integral, so you can't substitute $du = \frac{1}{2\sqrt{x}}dx$ 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:

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

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

$2\int e^u \, du$

**HallsofIvy** · February 26th 2011, 03:06 AM

Originally Posted by kikiya

Please help me evaluate

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

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

**kikiya** · February 26th 2011, 08:05 AM

Oh, I see the error! Thank you!

**kikiya** · February 26th 2011, 08:07 AM

Thank you for walking me through it! I was confiused about the balancing out, but now I get it! Thank you!

Thread: Evaluating an Integral

LinkBack

Thread Tools

Display

Evaluating an Integral

Similar Math Help Forum Discussions

Evaluating an integral

Help evaluating integral

Evaluating this integral

help evaluating an integral

Evaluating a double integral to find its signle integral.

Search Tags