## integration of exponential functions

I have a test tomorrow and I am hoping someone can help me with the following two integrals:

1. $$\int$$ye^$$^{-y^{2}}$$/2dy

For this one I am using substition with u=-y$$^{2}$$/2 and du = -y/4

What I don't understand is how we account for the first y in this integral.

2. $$\int$$(1/$${\sqrt{2}$$)e^$$^{-1/6(y-1)^{2}}$$dy

If I'm having trouble with the first one, you can assume I'm not getting far on this one either.

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 you're problem is unclear. can you re-type that plz.
 sure. one moment...

## integration of exponential functions

 Quote by Snarf du = -y/4
That is not right. Check the derivative of the y2 and then find du/dy.

 2. $$\int$$(1/$${\sqrt{2}$$)e^$$^{-1/6(y-1)^{2}}$$dy
This one's a bit unclear. Is it $$\exp\left(\frac{-1}{6}(y-1)^2\right)$$ or $$\exp\left(\frac{-1}{6(y-1)^2}\right)$$? Also should there be another y term in the integral somewhere?

 Recognitions: Gold Member Homework Help The first one is probably a lot easier than you think. When I see integrals like these, I always solve the derivative of a the exponential part and see what it looks like. What's the derivative of e^(-y^2)?

 Quote by neutrino That is not right. Check the derivative of the y2 and then find du/dy. This one's a bit unclear. Is it $$\exp\left(\frac{-1}{6}(y-1)^2\right)$$ or $$\exp\left(\frac{-1}{6(y-1)^2}\right)$$? Also should there be another y term in the integral somewhere?
The first one $$\exp\left(\frac{-1}{6}(y-1)^2\right)$$

 Quote by JasonRox The first one is probably a lot easier than you think. When I see integrals like these, I always solve the derivative of a the exponential part and see what it looks like. What's the derivative of e^(-y^2)?
-2ye^(-y^2) ?

 Quote by Snarf The first one $$\exp\left(\frac{-1}{6}(y-1)^2\right)$$
I don't think it's possible to solve the second integral in terms of elementary functions.

 neutrino the derivative of -y^2/2 would be -y?

 Quote by neutrino I don't think it's possible to solve the second integral in terms of elementary functions.
Would it help if it was being integrated from negative infinity to infinity?

Should have put that up initially

 So... for the first one would the answer be -2ye^(-y/2)?

 Quote by Snarf Would it help if it was being integrated from negative infinity to infinity? Should have put that up initially
Ah. There are some ways of dealing with that.

 or is it -2ye^(y^2)?

 Quote by Snarf So... for the first one would the answer be -2ye^(-y/2)?
 Quote by Snarf or is it -2ye^(y^2)?
Neither. The derivative of $$e^{-\frac{y^2}{2}}$$, with respect to y, $$-ye^{-\frac{y^2}{2}}$$.

 Quote by neutrino Neither. The derivative of $$e^{-\frac{y^2}{2}}$$, with respect to y, $$-ye^{-\frac{y^2}{2}}$$.
Oh. Now I see why you gave me that great hint.

Now as far as your hint on the second one... very interesting, but a bit complex for me. The last time I took a calc class was about three years ago.

Does this look like I'm getting close?

$$=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(1/6(y-1)^2+1/6(x-1)^2)}} dxdy$$

 You could use a sub. like u = (1/6)(y-1) to simplify things.