Question about change of variables

Chenkel said:

I was trying to draw a diagram on some paper to see what is going on with the Riemann sums with a change of variables, but it's a little tricky for me to understand.

PeroK said:

The proof of the change of variables uses previously established properties of the integral - in particular the fundamental theorems of calculus.

It would be difficult to go right back to first principles for something like that.

Chenkel said:

Hello everyone, ...

I'm just curious if I did these steps right and why the new integral in terms of theta gives the answer to the integral in terms of x appropriately.

Thank you!

SammyS said:

Yes.

You did those steps correctly and convincingly .

mathwonk said:

change of variables seems to me to be based on the definition of the derivative, i.e. the equation: [x(t)-x(t0)]/(t-t0) ≈ x'(t0). Thus in a Riemann sum, one replaces terms like [x(t)-x(t0)] by terms like x'(t0).[t-t0). or, in Leibniz notation, replace dx by (dx/dt).dt.

Chenkel said:

If it's hard to explain do not feel you have to, I do have the internet and can research the proof for the change of variables. But thank you for letting me know what the proof entails.

PeroK said:

The change of variables is really the inverse of the chain rule. Formally, if ##f## and ##u## are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$The proof is relatively simple:

Let ##F(y) = \int_0^y f(t)dt##. By the FTC (Fundamental Theorem of Calculus), ##F'(y) = f(y)##.

Let ##g(x) = F(u(x))##. By the chain rule: ##g'(x) = F'(u(x))u'(x) = f(u(x))u'(x)##.

Hence:
$$\int_a^bf(u(x))u'(x) \ dx = \int_a^bg'(x) \ dx = g(b) - g(a) = F(u(b)) - F(u(a))$$$$ = \int_{0}^{u(b)}f(t) \ dt - \int_{0}^{u(a)}f(t) \ dt = \int_{u(a)}^{u(b)}f(t) \ dt$$Note that it possible to use this result as it is stated. You should be on the look out for a situation where you have a composite function and the required derivative. For example:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx$$In this case, with ##f(x) = \frac{1}{\sqrt{x}}##, ##u(x) = 1+x^2## and ##u'(x) = 2x##, we see that we have a candidate for a change of variables:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx = \int_0^1f(u(x))u'(x) \ dx$$$$= \int_{u(0)}^{u(1)} f(t) \ dt = \int_1^2 \frac{1}{\sqrt{t}} \ dt = \bigg [-\frac 2 3 t^{-3/2} \bigg ]_1^2$$However, the change of variables is usually implemented using what is known as a substitution. In most cases, we start with a composite function in the integrand and make a substitution. For example, if we start with
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx$$Then we let ##x = \sin \theta## and ##dx = \cos \theta d\theta##. Then the substitution technique gives:
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx = \int_0^{\pi/2} \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \ d\theta$$We can see that the justification for this substitution technique is the change of variables theorem above. To see this, let's change the dummy variables here and rewrite this as:
$$\int_{0}^{\pi/2} \frac{\cos x}{\sqrt{1 - \sin^2 x}} \ dx = \int_0^1\frac 1{\sqrt{1 - t^2}} \ dt$$Now we can see that the change of variable applies with: ##f(x) = \frac 1 {\sqrt{1 - x^2}}##, ##u(x) = \sin x##, ##u'(x) = \cos x##, ##a = 0, b = \frac \pi 2##.

The vast majority of students I would say simply learn the substitution technique, and would struggle to justify it formally. Most would not be able to state the change of variable theorem.

Chenkel said:

Isn't the antiderivatie of ##\frac {1}{\sqrt{t}}## equal to ##2\sqrt{t}## not ##\frac {-2} {3} t^{\frac {-3}{2}}##?

Chenkel said:

Isn't the antiderivatie of ##\frac {1}{\sqrt{t}}## equal to ##2\sqrt{t}## not ##\frac {-2} {3} t^{\frac {-3}{2}}##?

PeroK said:

The change of variables is really the inverse of the chain rule. Formally, if ##f## and ##u## are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$The proof is relatively simple:

Let ##F(y) = \int_0^y f(t)dt##. By the FTC (Fundamental Theorem of Calculus), ##F'(y) = f(y)##.

Let ##g(x) = F(u(x))##. By the chain rule: ##g'(x) = F'(u(x))u'(x) = f(u(x))u'(x)##.

Hence:
$$\int_a^bf(u(x))u'(x) \ dx = \int_a^bg'(x) \ dx = g(b) - g(a) = F(u(b)) - F(u(a))$$$$ = \int_{0}^{u(b)}f(t) \ dt - \int_{0}^{u(a)}f(t) \ dt = \int_{u(a)}^{u(b)}f(t) \ dt$$Note that it possible to use this result as it is stated. You should be on the look out for a situation where you have a composite function and the required derivative. For example:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx$$In this case, with ##f(x) = \frac{1}{\sqrt{x}}##, ##u(x) = 1+x^2## and ##u'(x) = 2x##, we see that we have a candidate for a change of variables:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx = \int_0^1f(u(x))u'(x) \ dx$$$$= \int_{u(0)}^{u(1)} f(t) \ dt = \int_1^2 \frac{1}{\sqrt{t}} \ dt = \bigg [2 t^{1/2} \bigg ]_1^2$$However, the change of variables is usually implemented using what is known as a substitution. In most cases, we start with a composite function in the integrand and make a substitution. For example, if we start with
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx$$Then we let ##x = \sin \theta## and ##dx = \cos \theta d\theta##. Then the substitution technique gives:
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx = \int_0^{\pi/2} \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \ d\theta$$We can see that the justification for this substitution technique is the change of variables theorem above. To see this, let's change the dummy variables here and rewrite this as:
$$\int_{0}^{\pi/2} \frac{\cos x}{\sqrt{1 - \sin^2 x}} \ dx = \int_0^1\frac 1{\sqrt{1 - t^2}} \ dt$$Now we can see that the change of variable applies with: ##f(x) = \frac 1 {\sqrt{1 - x^2}}##, ##u(x) = \sin x##, ##u'(x) = \cos x##, ##a = 0, b = \frac \pi 2##.

The vast majority of students I would say simply learn the substitution technique, and would struggle to justify it formally. Most would not be able to state the change of variable theorem.

Chenkel said:

So just to prove my formulas are consistent with the change of variables formula you proved.

##f(x) = \sqrt{r^2 - x^2}##
##u(\theta) = r\cos(\theta)##
##u'(\theta) = -r\sin(\theta)##

If you expand the following equation with the previous variables it's the same equation I used in my proof
##\int_\pi^0 f(u(\theta))u'(\theta)d\theta = \int_{-r}^{r} f(x)dx##
Which is the same as
##\int_\pi^0 f(u(\theta))u'(\theta)d\theta = \int_{u(\pi)}^{u(0)} f(t)dt##
Which seems to be consistent with the change of variables formula you proved.

Does this justify the change of variables in my proof?

PeroK said:

The change of variables is really the inverse of the chain rule. Formally, if ##f## and ##u## are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$The proof is relatively simple:

Let ##F(y) = \int_0^y f(t)dt##. By the FTC (Fundamental Theorem of Calculus), ##F'(y) = f(y)##.

Let ##g(x) = F(u(x))##. By the chain rule: ##g'(x) = F'(u(x))u'(x) = f(u(x))u'(x)##.

Hence:
$$\int_a^bf(u(x))u'(x) \ dx = \int_a^bg'(x) \ dx = g(b) - g(a) = F(u(b)) - F(u(a))$$$$ = \int_{0}^{u(b)}f(t) \ dt - \int_{0}^{u(a)}f(t) \ dt = \int_{u(a)}^{u(b)}f(t) \ dt$$Note that it possible to use this result as it is stated. You should be on the look out for a situation where you have a composite function and the required derivative. For example:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx$$In this case, with ##f(x) = \frac{1}{\sqrt{x}}##, ##u(x) = 1+x^2## and ##u'(x) = 2x##, we see that we have a candidate for a change of variables:
$$\int_0^1 \frac{2x}{\sqrt{ 1 + x^2}} \ dx = \int_0^1f(u(x))u'(x) \ dx$$$$= \int_{u(0)}^{u(1)} f(t) \ dt = \int_1^2 \frac{1}{\sqrt{t}} \ dt = \bigg [2 t^{1/2} \bigg ]_1^2$$However, the change of variables is usually implemented using what is known as a substitution. In most cases, we start with a composite function in the integrand and make a substitution. For example, if we start with
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx$$Then we let ##x = \sin \theta## and ##dx = \cos \theta d\theta##. Then the substitution technique gives:
$$\int_0^1\frac 1{\sqrt{1 - x^2}} \ dx = \int_0^{\pi/2} \frac{\cos \theta}{\sqrt{1 - \sin^2 \theta}} \ d\theta$$We can see that the justification for this substitution technique is the change of variables theorem above. To see this, let's change the dummy variables here and rewrite this as:
$$\int_{0}^{\pi/2} \frac{\cos x}{\sqrt{1 - \sin^2 x}} \ dx = \int_0^1\frac 1{\sqrt{1 - t^2}} \ dt$$Now we can see that the change of variable applies with: ##f(x) = \frac 1 {\sqrt{1 - x^2}}##, ##u(x) = \sin x##, ##u'(x) = \cos x##, ##a = 0, b = \frac \pi 2##.

The vast majority of students I would say simply learn the substitution technique, and would struggle to justify it formally. Most would not be able to state the change of variable theorem.

Chenkel said:

What do we do if ##a## and ##b## are largely separated and ##u## is a periodic function that oscillates more than once in the interval ##[a, b]##?

For example I chose ##u=r\cos(\theta)## and I chose ##\theta = [\pi, 0]## and I got the correct answer for the area of the circle, but if I chose ##\theta = [3\pi, 0]## I would have gotten the wrong result for the area of the circle with ##[a=3\pi, b=0]##, if I chose the latter my result for the area of the circle would have been ##3\pi{r^2}##

Any tips or keys to understand how to handle the substitution rule on large intervals with periodic functions?

Thanks again for teaching me.

PeroK said:

The change of variables is really the inverse of the chain rule. Formally, if ##f## and ##u## are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$

PeroK said:

You do have to be careful. Normally, the substitution is done the other way round, with a different meaning for ##u##. In that case, ##u(x)## must be an invertible function.

Here is the change of variables:

And here is a substitution, with ##x = u(\theta)## and ##dx = u'(\theta) d \theta##.
$$\int_a^bf(x) \ dx = \int_{u^{-1}(a)}^{u^{-1}(b)}f(u(\theta)) u'(\theta) \ d\theta$$Note that I've distinguished between variables ##x, \theta## and functions ##f, u##. And, you can see explicitly that ##u## must be one-to-one and invertible.

Chenkel said:

So perhaps the reason my proof worked is because when I let ##u=r\cos(\theta)##
I kept ##\theta## in the interval ##[0, \pi]## and in that interval u is one-to-one and and invertible if I'm not mistaken.

Chenkel said:

Now I'm a little confused and not sure of I said that right

Chenkel said:

So perhaps the reason my proof worked is because when I let ##u=r\cos(\theta)##
I kept ##\theta## in the interval ##[0, \pi]## and in that interval u is one-to-one and and invertible if I'm not mistaken.

PeroK said:

Yes. Here's how it can go wrong. With ##x = \sin \theta##, we have:
$$\int_0^{2\pi} \sin \theta \ d\theta = \int_0^0 \dots$$And it's a mess.

PeroK said:

Yes. Here's how it can go wrong. With ##x = \sin \theta##, we have:
$$\int_0^{2\pi} \sin \theta \ d\theta = \int_0^0 \dots$$And it's a mess.

PeroK said:

I'm a little surprised that you don't see the general problem. It's the null interval. Although, perhaps it wasn't the best example. Try this instead:

With ##x = \sin \theta##, we have:
$$\int_0^{2\pi} \sin^2 \theta \ d\theta = \int_0^0 \dots$$And, in fact, every integral on the interval ##[0, 2\pi]## has the same null result.

PeroK said:

I'm a little surprised that you don't see the general problem. It's the null interval. Although, perhaps it wasn't the best example. Try this instead:

With ##x = \sin \theta##, we have:
$$\int_0^{2\pi} \sin^2 \theta \ d\theta = \int_0^0 \dots$$And, in fact, every integral on the interval ##[0, 2\pi]## has the same null result.

Chenkel said:

The fact that the proof you provided for the change of variables definite integration equation doesn't work with a non-injective ##u## function doesn't diminish the validity of the proof itself, right?

PeroK said:

As I've pointed out several times(!), the proof uses the chain rule and goes in the opposite direction from the usual change of variables technique. There is no problem with the function ##u(x)## in the proof not being invertible. You could try it out.

Chenkel said:

So ##\int_a^b f(u(x))u'(x)dx = \int_{u(a)}^{u(b)} f(t)dt## is still true when ##u(a) = u(b)##?