Integration of exponential functions

[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.

 

[rocomath]

you're problem is unclear.

can you re-type that please.

 

[Snarf]

sure. one moment...

 

[neutrino]

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?

 

[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)?

 

[Snarf]

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

 

[Snarf]

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) ?

 

[neutrino]

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.

 

[Snarf]

neutrino

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

 

[Snarf]

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

 

[Snarf]

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

 

[neutrino]

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.
For example, https://www.physicsforums.com/showthread.php?t=71285

 

[Snarf]

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

 

[neutrino]

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

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

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

 

[Snarf]

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?

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

 

[neutrino]

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