# Question about change of variables

• B
• Chenkel
Chenkel
Hello everyone,

I found a good proof for the area of a circle being ##{\pi}r^2## but I was recently working on my own proof and I used a change of variables and was wondering if I did it correctly and why a change of variables seems to work.

I start with the equation of a circle ##r^2 = x^2 + y^2## then I set up the integral ##2\int_{-r}^{r} {\sqrt{(r^2 - x^2)}}{dx}##

Now I change variables with ##x=r\cos(\theta)## and ##dx=-r\sin(\theta)d{\theta}##

And I update the integral
##-2\int_{\pi}^{0} {\sqrt{(r^2 - (r\cos(\theta))^2)}}r\sin(\theta){d\theta}##
Which is the same as
##2\int_{0}^{\pi} \sqrt{(r^2 - (r\cos(\theta))^2)}r\sin(\theta){d\theta}##
And using trig identities this is the same as
##2\int_{0}^{\pi} {{r^2}{(\sin(\theta))^2}}{d\theta}##
Finally using the half angle trig identity this is the same as ##{r^2}\int_{0}^{\pi} {(1 - \cos(2\theta))}{d\theta}##
The half angle identity is ##(sin(\theta))^2 = (\frac 1 2)(1 - \cos(2\theta))##
The antiderivatie is ##\theta - \frac {\sin(2\theta)}{2}## the evaluation of the full integral gives ##{\pi}r^2##

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.

Any help is appreciated,

Thank you!

Last edited:
PeroK
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.

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.
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
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.
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.

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!
Yes.

You did those steps correctly and convincingly .

Chenkel
SammyS said:
Yes.

You did those steps correctly and convincingly .
Thanks for the feedback!

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
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.
I'm still learning about it myself, the Riemann sum seems to have some analogs to differentials as the sample size approaches infinity. It seems you might be hinting at this in your last reply.

So I believe I've been grocking the integration process more, in the first integral we are summing vertical strips with respect to x and then when we change the variable of integration to ##\theta## by reexpressing the differential area with a change in variables for the differential element with new limits, this transformation of the terms of the differential element into a different expression of the same thing does not make the integral invalid, it's infinitesimal analysis and transformations of infinitesimals into new valid expressions should be permitted.

Last edited:
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.
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.

Last edited:
docnet, pinball1970 and Chenkel
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.
Isn't the antiderivatie of ##\frac {1}{\sqrt{t}}## equal to ##2\sqrt{t}## not ##\frac {-2} {3} t^{\frac {-3}{2}}##?

Thanks again for the result of the proof, I'm going to study it some more today but I think I mostly understood it.

PeroK
Chenkel said:
Isn't the antiderivatie of ##\frac {1}{\sqrt{t}}## equal to ##2\sqrt{t}## not ##\frac {-2} {3} t^{\frac {-3}{2}}##?
Yes, of course.

Chenkel
Chenkel said:
Isn't the antiderivatie of ##\frac {1}{\sqrt{t}}## equal to ##2\sqrt{t}## not ##\frac {-2} {3} t^{\frac {-3}{2}}##?
Yes! Fixed.

Chenkel
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.

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?

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?
Yes, you applied the technique correctly. The point of my post was to show that it comes ultimately from the chain rule.

pinball1970, SammyS and Chenkel
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.
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.

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.
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:
PeroK said:
The change of variables is really the inverse of the chain rule. Formally, if ##f## and ##u## are suitable functions:
$$\int_a^bf(u(x))u'(x) \ dx = \int_{u(a)}^{u(b)}f(t) \ dt$$
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
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.
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:
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.
Now I'm a little confused and not sure of I said that right

Chenkel said:
Now I'm a little confused and not sure of I said that right
I think I probably said it right, I hope so.

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.
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.

Chenkel
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.
Interesting, thank you for the insight!

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.
I'm not sure what you mean by somethings going wrong there.

The integral ##\int_0^{2\pi} \sin \theta \ d\theta## is 0 using the fundamental theorem of calculus ##[-cos(\theta)]\Bigg|_0^{2\pi}=-\cos(2\pi) + cos(0) = 0## and the right hand side of the equation you posted is the integral from ##0## to ##0## of some function which is always zero.

I believe ##u## has to be one-to-one and if it's one-to-one it's also invertible but I'm not exactly sure of the reasoning behind this requirement for the change of variable theorem to work.

Sorry if I'm getting too into the weeds, I just want to understand precisely.

And thanks again for the help!

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
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.
I agree that the left hand side of that equation doesn't equal the right side, and I think I see the general problem, if ##u=\sin \theta## and the limits of integration are ##0## to ##2\pi## then the result will be null in the change of variables formula regardless of what the original function is.

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.
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?

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?
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
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.
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)##?

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)##?
Yes. That's a valid scenario. In that case, the integral on the left-hand side will always be zero. You can look at the proof above and see how it works in the general case, but as an example:

Let ##a = 0, b = \pi, u(x) = sin(x), f(x) = x^2##:
$$\int_0^{\pi} \sin^2(x)cos(x) \ dx = \bigg [\frac 1 3 \sin^3 x \bigg ]_0^{\pi} = 0$$

Chenkel

• Calculus
Replies
2
Views
375
• Calculus
Replies
4
Views
475
• Calculus
Replies
8
Views
298
• Calculus
Replies
3
Views
1K
• Calculus
Replies
12
Views
2K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
3
Views
739
• Calculus
Replies
16
Views
1K
• Calculus
Replies
4
Views
1K
• Calculus
Replies
2
Views
2K