# I Integration - chain rule / functional

1. Oct 19, 2016

### binbagsss

I have $\int_{t = 0}^{t = 1} \frac{1}{x} \frac{dx}{dt} dt = \int_{t = 0}^{t = 1} (1-y) dt$ [1]
The LHS evaluates to $ln \frac{(x(t_0+T))}{x(t_0)}$, where $t_{1}=t_{0}+T$

My issue is that, asked to write out the intermediatary step, I could not. I am unsure how you do this when the limits aren't expressed in terms of $x$ here. So I can see that $1/x dx = ln x$, but I'm unsure of what has happened to $'\frac {dt}{dt}'$ and how the integration limits are done properly.

I think the two below points tie in, and link to where my understanding is lacking with this:

1) Am I correct in thinking that, given $dx/dt = x(1-y)$, in order to get to expression [1] you can not simply, formally, 'seperate variables' - see point below, but rather you should divide by $x$ and then integrate both sides w.r.t $t$.

2) From high school up to now, I have used seperation of variables to solve such things as : $dx/dt=b(x)$ => $dx/b(x)=dt$

However my lecturer today told me that is inproper and rather one should use a integration factor - i think understanding this point ties with/ may help my initial question?

Last edited by a moderator: Oct 19, 2016
2. Oct 19, 2016

### Staff: Mentor

Without knowing the relationship between x and t, I don't know how one would evaluate the integral on the left, since it isn't known what x(1) and x(0) are.
Well, the above should be $\int 1/x dx = \ln|x| + C$

3. Oct 19, 2016

### binbagsss

Sorry, I missed the constant C, yes.

my question wasn't about evaluating x(1) and x(0)? it was about how you get from line 1 to line 2 and 'transform' from the limits in $t$ to $x$ limits, I'm only ever used to seeing $x$ limits in such an expression... (evaluates was probably the wrong word...)

4. Oct 19, 2016

### Staff: Mentor

You can't transform the t limits to x limits unless you know how x and t relate; that is, unless you know a formula for x(t).

5. Oct 19, 2016

### binbagsss

I mean how its gone to $x(t_{1})$ and $x(t_{0}$), not explicitly what they are.

The only other information that I have not included in the OP is that $T$ is the period of the solution so the LHS evaluates to $0$ ...
$x$ and $t$ relate by the equation in 1) in the OP and we are also given $dy/dt=K(-x+xy)$

6. Oct 19, 2016

### Staff: Mentor

When you change the variable of integration, by for example, using a substitution, you usually need to change the limits of integration. The exception is when you "undo" the substitution after finding an antiderivative.

Here's a very simple example,
$$\int_{x = 0}^1 (x + 2)^3 dx$$
Let u = x + 2
Then du = dx
And x = 0 => u = 2, x = 1 => u =3
Carrying out the substitution and changing the limits of integration:
$$\int_{x = 0}^1 (x + 2)^3 dx = \int_{u = 2}^3 u^3 du \\ = \frac{u^4}{4}|_2^3 = \frac{81}{4} - \frac{16}{4} = \frac{65} 4$$
I don't see how equation 1 shows a specific relationship between x and t.

7. Oct 19, 2016

### binbagsss

yeh that example is fine. but i'm still not understanding what has happened in my initial question...

8. Oct 19, 2016

### binbagsss

Apologies in the OP the limits over the intergral should read as $^{t_{1}} _{t_{0}} \int$, where $t_{1}=t_{0}+T$

9. Oct 19, 2016

### Staff: Mentor

It's customary to put the integration limits to the right of the integral sign, like so:
$\int_{t_0}^{t_1} f(t) dt$

10. Oct 19, 2016

### binbagsss

okay. i was concerning the change in limits, as not being explicit..
still no idea how to approach my initial question.

11. Oct 19, 2016

### Staff: Mentor

In my example there is a clear relationship between u and x; namely, u(x) = x + 2.
In your integral, there is no such relationship, at least one that I can see.

12. Oct 19, 2016

### binbagsss

I agree.
my OP is what we done in lectures however.

13. Oct 19, 2016

### Stephen Tashi

Can you quote the statements that came before the equation in your OP ?

14. Oct 21, 2016

### binbagsss

Literally all information has been posted above. But I shall summarise again:

System of equations:

$dx/dt=x-xy$ [1]
$dy/dt= K(-y+xy)$ [2], $K$ a constant.

The lecturer makes the statement that many of the solutions of $x(t)$ and $y(t)$ are periodic.
Let the period be $T$.

My lecturer then jumps to the line:

$\int _{t_{0}}^{t_{0}+T} 1/x (dx/dt) dt = \ln (\frac{x(t_{0}+T)}{(x(t_{0})})=0= \int _{t_{0}} ^{t_{0}+T} (1-Ky) dt$

Last edited by a moderator: Oct 21, 2016
15. Oct 21, 2016

### Staff: Mentor

The integral on the left could be rewritten as $\int_{x(t_0)}^{x(t_0 + T)} \frac {dx} x$.
Since x(t) is assumed to be periodic with period T, then this integral is 0, since $x(t_0) = x(t_0 + T)$.
However, saying that both x(t) and y(t) are periodic doesn't necessarily imply that both functions have the same period.

Last edited: Oct 21, 2016
16. Oct 21, 2016

### binbagsss

Last edited by a moderator: Oct 21, 2016
17. Oct 21, 2016

### Staff: Mentor

Hints about what? I explained why the integral on the left (i.e., $\int_{x(t_0)}^{x(t_0 + T)} \frac {dx} x$) is zero. As I already said $x(t_0) = x(t_0 + T)$, assuming that x(t) is periodic with period T, Then $\frac {x(t_0 + T)}{x(t_0)} = 1$, so ln of this fraction is zero.
If both x(t) and y(t) have the same period, then $\int_{x(t_0)}^{x(t_0 + T)} ydt$ would also be zero, and I leave it as an exercise for you to explain why $\int_{x(t_0)}^{x(t_0 + T)} (1 - Ky)dt$ is also zero.

18. Oct 22, 2016

### binbagsss

My initial question.
I know why the integral evaluates to 0.
and this wasn't my question
it says 'asked to show the intermediatary step, I could not'

19. Oct 22, 2016

### Staff: Mentor

What do you mean by the intermediate step? ("intermediatary" is not a word...)

20. Oct 22, 2016

### binbagsss

why are you bothering to reply if your not addressing the question? you're sharing your general knowledge on other parts of the question...