# Integration By Substitution and Line Integrals

1. Mar 12, 2004

### Claire84

For our homework this week for Pure, one of the questions is to ingration 1/(16 + x^2) with respect to x between the limits 0 and 4. I know the result from the formula wqith arctan in it, but since we've to use substitituion here and not just plonk down the formula, I'm confused as to what to substitute. Do I need to rearrange this eqt to work it out? I've got in my Applied notes how to differentiate the arctan thing to get this, but it doesn't make sense as to what I've to substitue in because at least when you differentiate you can work with eqt better. Even a hint to set me on the right track would be much-appreciated as I've got the other substituion questions okay but this one is bugging me as there seems to be more to it than the others.

Just one other thin about line integrals that we covered today. They make sense until I see what you've to put into the limits. For example
I= integral from A to B of xydx + integral from A to B of y^2dy

where A=(0,0) and B=(1,2)

The first case is where x=t and y=2t^2 and in brackets beside it has 0<=t<=2

I'vwe done the question and used he limits as 0 and 2, but only because it says beside the question. not because I know why! I mean whyy are the points given then? Do you work the limits out for t from the x=t and y=2t^2 eqts?

2. Mar 12, 2004

### HallsofIvy

Staff Emeritus
To integrate dx/(16 + x^2): Since you already know that the integral of dx/(1+x^2) is arctan(x)+ C, the problem is that "16". If the problem were dy/(16+ 16y^2) then you could write it as
(1/16)(dy/(1+y^2)) and integrate that. Okay, how about x=4y (from
x^2= 16y^2)?

How about looking at it the other way? You are given that the path is x= t and y= 2t^2. When t= 0, what are x and y? When t= 1, what are x and y?

Yes, you work the limits out from x= t and y= 2t^2. You are to integrate "from (0,0) to (1,2)". That is: for the starting point
x= t= 0 and y= 2t^2= 0. Yes, those are true only for t= 0. For the endpoint, x= t= 1 and 2t^2= 2. Again, those are true only for t= 1.

3. Mar 13, 2004

### Claire84

Okay, so I'm only integrating between 0 and 1 then? That's okay, I was confused because beside that question he had put between 0 and 2.

For the second part we have the same points but x=2t^3 and y=t^2 I can't find the same t that works for these apart from 0, so I'm not sure what is to go as my upper limit. Beside this one we have in brackets (0<=t<=1)

4. Mar 13, 2004

### Claire84

Could it be an error in the question? I'm also able to work out the limits from the examples in our notes but for this one (given in the post above) I can't find the uper limit at all. If no one can see a logical answer to it I'll go speak to the lecturer on Monday, but please save me from that.

5. Mar 13, 2004

### matt grime

So the points between which we must find the integral are (0,0) and (1,2)

the parametrization

x=2t^3, y=t^2, does not describe a curve passing through those points. it passes through the point (2,1) at t=1, which strongly suggests a typo - perhaps the x and y got swapped in the typing up.

6. Mar 13, 2004

### Claire84

Thanks for that. I think that was the bit that was confusing me the most about how to work out the limits because I didn't know where the pregnant woman, sorry male lecturer, had got that from. Much-appreciated. Seems I'll have to mention it to the grumpy old bear on Monday, or maybe an email would be a less hairy experiebce. Just kdding, I'm sure he's delightful, really!