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Do yall remember your calculus??

Sgt.Rock

Grizzled Veteran
May 3, 2006
482
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Texas
I've got a final tomorrow and ive been working my butt of studying for it, but I've run into a snag.


It has to do with e's, and Ive forgotten it from high school calc classes

I know the int of e^x = e^x but.....



The int of e^(x^2) dx = (e^(x^2))/2x + C correct???

You set u=x^2 du=2x dx THEN du/2x = dx


Is that right??



The same goes for int of e^(2x) = (e^(2x)/2) + C

and

int of e^(x/2) = 2e^(x/2) + C ?????


Please Help! :(




P.S. What is the derivitive to the first one??
 
I'll start from the bottom because its relevant. Also, its been a few months since calc class, so my notation is most likely way off.
P.S. What is the derivitive to the first one??
X^2 is a function, so you need to use the derivative for e^(u), where u is a function of x. Your derivative is dy/dx= ln(u)e^(u) x du/dx.

The int of e^(x^2) dx = (e^(x^2))/2x + C correct???

You set u=x^2 du=2x dx THEN du/2x = dx


Is that right??

Nope, when doing the "u substitution" method you cannot have a variable materialize during the substitution (in your case the 1/2x). One way you can solve this problem is integration by part as shown below, or even an easier method shown way below:
Integration by part:
Set u = E^(x^2)
then du= ln(x^2) (E^(x^2)) (2x)

Your dv then = dx
and v = x

Then you do your integration by part, using the following formula:
uv - INT [(v)(du)]
and keep going till you win. This equation will very likely turn very long and ugly so its not the best method, although the skill is extremely useful in itself.

/edit Remembered something else. This is is a much easier method:
If you take a natural log of both sides, you end up with
ln(y) = x^2 ln (e)
ln(y) = x^2

Now, using the integration by part, you integrate both sides and get (hopefully i got this right):
y (ln(y)-1) = (x^3)/3 + C
Since we know that ln(y) = x^2, we can rewrite the above as
y (x^2-1) = (x^3)/3+C and simplify to get the answer. Way way easier than the integration by part, but it was needed here so nyah.


The same goes for int of e^(2x) = (e^(2x)/2) + C

and

int of e^(x/2) = 2e^(x/2) + C ?????
In this case you use u substitution, since no new variables will be created by doing so. So you are correct.



.....I think. :eek:
 
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Oi, where did the ln come from for the derivitive?? I have finally found the derivitave of e^(x^2) (or the way described on how to do it) in my guide book, and they say (like you said), to work it as a function of x

So....

d/dx e^(u) with u being the function of x (which is x^2) so that the derivative is equal to:

(e^(u))(u')

Which ends up giving me (2x)(e^(x^2)).


But......


I got ahold of my Asian friend :)D:D:p) with a Ti-89, and according to him and the calc, the Integral of e^(x^2) doesnt exist. Then I checked my problems with it in it, and the were all either divided by that e function or multiplied by it, causing substition by u to work perfectly.

Args, but thank you for the reply, it made me think harder (especially getting certain functions of x with derivatives to be u).


False alarm on my part :( :p
 
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