# Integration of exponential functions

1. Oct 28, 2007

### Snarf

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.

2. Oct 28, 2007

### rocomath

you're problem is unclear.

can you re-type that plz.

3. Oct 28, 2007

### Snarf

sure. one moment...

4. Oct 28, 2007

### 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?

5. Oct 28, 2007

### 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)?

6. Oct 28, 2007

### Snarf

The first one $$\exp\left(\frac{-1}{6}(y-1)^2\right)$$

7. Oct 28, 2007

### Snarf

-2ye^(-y^2) ?

8. Oct 28, 2007

### neutrino

I don't think it's possible to solve the second integral in terms of elementary functions.

9. Oct 28, 2007

### Snarf

neutrino

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

10. Oct 28, 2007

### Snarf

Would it help if it was being integrated from negative infinity to infinity?

Should have put that up initially

11. Oct 28, 2007

### Snarf

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

12. Oct 28, 2007

### neutrino

13. Oct 28, 2007

### Snarf

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

14. Oct 28, 2007

### neutrino

Neither. The derivative of $$e^{-\frac{y^2}{2}}$$, with respect to y, $$-ye^{-\frac{y^2}{2}}$$.

15. Oct 28, 2007

### Snarf

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?

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

16. Oct 28, 2007

### neutrino

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