# Homework Help: Integration problem am I correct?

1. May 11, 2010

1. The problem statement, all variables and given/known data
Integrate: $$\int$$2x*ex2dx

2. Relevant equations
$$\int$$eudu = eu

3. The attempt at a solution
I tried integration by substitution:
Let u=x2
Then du=2x*dx

So then it is:
$$\int$$eu*du

Which would give me:
eu+c

So subbing x2 back in for u:
ex2+c

I looked this type of function up online and got a lot of complicated results involving square root of pi or a log function and things referring to an error formula integral or something? ... so I was just double checking to see if this seemingly simple solution works. And maybe someone could explain what I was seeing online? Thanks

Last edited: May 11, 2010
2. May 11, 2010

### physicsman2

Well first you forgot the dx in the original expression.

Your work and answer looks right to me. I wouldn't look online for ways to do things or answers because you usually will either find incorrect things or stuff that isn't at your level yet.

3. May 11, 2010

### LCKurtz

4. May 11, 2010

Thanks, yeah I remember that now. Worked it out back and forth now. Thanks for the help! It's been a while since I did calc out like this so I keep forgetting my little tips and tricks.

5. May 11, 2010

### Char. Limit

Also, badgerman, what you solved (the integral of 2x*e^(x^2)) and what you looked up online (the integral of e^(-x^2)) are different functions and that's why what you looked up was strange. I'm guessing it was something like this?

$$\integral e^{-x^2} dx = \frac{\sqrt{\pi}}{2} erf(x) + C$$

That's the integral of a different, albeit similar function. Just to clear up what you DID see.

6. May 11, 2010

Yeah I noticed the negative sign when I was googling it. Amazing how a single negative can change it so much. I think also a lot of the confusion arose from finding a definite as opposed to indefinite integral, in terms of an exact numerical solution. Either way, it's all coming back to me now... once again, thanks for the help :)

7. May 11, 2010

### Cyosis

No it is not the negative sign that makes the difference, but the x in front of e^x^2. As you can see by differentiation e^x^2 we obtain 2xe^x^2. There is however no 'standard' function that when differentiated gives e^x^2.

8. May 11, 2010