Geometric Derivation of f'(x) = arctan(x) for f(x)=∫(^x, _0) (dt)/(1+t^2)

[Laurentiusson]

Show that the derivative of the integral by looking at it geometrically and using the squeeze theorem:

f(x) = ∫(^x, _0) (dt)/(1+t^2)

is equal to arctan(x).

Clarifying picture: http://i.imgur.com/OgprJ1Z.png

I have done this algebraic but I'm struggling to do it with the conditions given. Sadly I don't find anything similar in my course book.

Someone know how to do this?

f(x)=∫(^x, _0) (dt)/(1+t^2)

Let g(t) = 1/(1+t^2)
Let h(x) = x
h'(x) = 1

f'(x) = g ( h(x) ) h'(x)
f'(x) = g(x) (1)
f'(x) = 1/(1+x^2)

The derivative of this integral is 1/(1+x^2)

How about the geometrical part of it?

 

[LCKurtz]

Show that the derivative of the integral by looking at it geometrically and using the squeeze theorem:

f(x) = ∫(^x, _0) (dt)/(1+t^2)

is equal to arctan(x).

But the derivative of that integral is not $\arctan x$, so that obviously is not the correct statement of the problem. Please give the exact statement of the problem.

 

[Laurentiusson]

But the derivative of that integral is not $\arctan x$, so that obviously is not the correct statement of the problem. Please give the exact statement of the problem.

That is the problem. And since it's originally written in Swedish I don't think it will help you anyway.

 

[LCKurtz]

That can't be the correct statement of the problem because it isn't true. I can guess what you are supposed to do:
Since $\arctan x = \int_0^x \frac 1 {1+t^2}~dt$, show geometrically that $\frac d {dx}\arctan x = \frac 1 {1+x^2}$. Assuming that is the problem I would suggest using the difference quotient limit definition of derivative on that integral followed by the mean value theorem for integrals. I guess that could be thought of as a geometric argument and it does use the squeeze principle.

 

[vela]

Well, you showed that the derivative of the integral is 1/(1+x^2), and you're being asked to show that that's equal to arctan x. Clearly, arctan x doesn't equal 1/(1+x^2), so there's no point in trying to do the problem as you stated.

 

[Laurentiusson]

That can't be the correct statement of the problem because it isn't true. I can guess what you are supposed to do:
Since $\arctan x = \int_0^x \frac 1 {1+t^2}~dt$, show geometrically that $\frac d {dx}\arctan x = \frac 1 {1+x^2}$. Assuming that is the problem I would suggest using the difference quotient limit definition of derivative on that integral followed by the mean value theorem for integrals. I guess that could be thought of as a geometric argument and it does use the squeeze principle.

How to do this?

 

[LCKurtz]

How to do this?

You have to show some effort. Use the definition of derivative:$$
f'(x) = \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$$on that integral and show us what happens.

 

[Laurentiusson]

You have to show some effort. Use the definition of derivative:$$
f'(x) = \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$$on that integral and show us what happens.

I'm not even sure how to start. My little messy atemptent is: (f((da)/(1+a^2)+h)−f(a))/((h))

 

[LCKurtz]

In your problem, $f(x) = \int_0^x \frac 1 {1+t^2}~dt$. Start with that.

 

[Laurentiusson]

In your problem, $f(x) = \int_0^x \frac 1 {1+t^2}~dt$. Start with that.

That derivate would be -((2t)/(1+t²)²

 

[LCKurtz]

Wild guesses don't count. If you and I are going to make any progress on this problem you need to look at posts #7 and #9 and try that.

 

[Laurentiusson]

Wild guesses don't count. If you and I are going to make any progress on this problem you need to look at posts #7 and #9 and try that.

It's not a wild guess. Please give fair criticism.

f(x)=1/1+t² dt

f(x+h)=1/1+(t+h)² dt

Is that the correct start?

 

[LCKurtz]

It's not a wild guess.

Without any work it looked like a wild guess to me.

Please give fair criticism.

f(x)=1/1+t² dt

f(x+h)=1/1+(t+h)² dt

Is that the correct start?

No. You have a function of $x$ in this problem. The $t$ is a dummy variable in the integral. And a function of $x$ is not a function of $t$.

To start on this problem you need to literally plug the function of $x$ in post #9 into the formula in post #7. You will have a fraction with a couple of integrals in it. That is what you have to work on.

 

[Mark44]

That derivate would be -((2t)/(1+t²)²

You are ignoring the fact that f(x) involves a definite integral.

Side note: The correct term is "derivative". To the best of my knowledge, there is no such word in English as "derivate".

 

[Laurentiusson]

Without any work it looked like a wild guess to me.No. You have a function of $x$ in this problem. The $t$ is a dummy variable in the integral. And a function of $x$ is not a function of $t$.

To start on this problem you need to literally plug the function of $x$ in post #9 into the formula in post #7. You will have a fraction with a couple of integrals in it. That is what you have to work on.

f(x) = ∫_0^x dt/(1+t^2)

f(x+h) = ∫_0^(x+h) dt/(1+t^2)

f(x+h) - f(x) = ∫_0^(x+h) dt/(1+t^2) - ∫_0^x dt/(1+t^2) = ∫_x^(x+h) dt/(1+t^2)

(f(x+h) - f(x))/h = (1/h) ∫_x^(x+h) dt/(1+t^2)

I guess?

 

[Mark44]

f(x) = ∫_0^x dt/(1+t^2)

f(x+h) = ∫_0^(x+h) dt/(1+t^2)

f(x+h) - f(x) = ∫_0^(x+h) dt/(1+t^2) - ∫_0^x dt/(1+t^2) = ∫_x^(x+h) dt/(1+t^2)

(f(x+h) - f(x))/h = (1/h) ∫_x^(x+h) dt/(1+t^2)

I guess?

So far, so good. A graph that represents the integral would be a good thing to make. Plus, you need to get a lower bound and an upper bound on the area the integral represents, and then use the squeeze theorem to find the limit of what you have above.

Your integrals would be easier to read if you were using LaTeX. You've already done most of the work.
# #f(x+h) - f(x) = \int_0^{x+h} dt/(1+t^2) - \int_0^x dt/(1+t^2) = \int_x^{x+h} dt/(1+t^2) # #
All I did was replace your integral signs with \int and change a couple of pairs of () to braces {}

Removing the spaces in the first and last pairs of # characters results in this:
$f(x+h) - f(x) = \int_0^{x+h} dt/(1+t^2) - \int_0^x dt/(1+t^2) = \int_x^{x+h} dt/(1+t^2)$

Here it is again using \frac to write fractions a little nicer.
$f(x+h) - f(x) = \int_0^{x+h} \frac{dt}{1+t^2} - \int_0^x \frac{dt}{1+t^2} = \int_x^{x+h} \frac{dt}{1+t^2}$

 

[Laurentiusson]

So far, so good. A graph that represents the integral would be a good thing to make. Plus, you need to get a lower bound and an upper bound on the area the integral represents, and then use the squeeze theorem to find the limit of what you have above.

Your integrals would be easier to read if you were using LaTeX. You've already done most of the work.
# #f(x+h) - f(x) = \int_0^{x+h} dt/(1+t^2) - \int_0^x dt/(1+t^2) = \int_x^{x+h} dt/(1+t^2) # #
All I did was replace your integral signs with \int and change a couple of pairs of () to braces {}

Removing the spaces in the first and last pairs of # characters results in this:
$f(x+h) - f(x) = \int_0^{x+h} dt/(1+t^2) - \int_0^x dt/(1+t^2) = \int_x^{x+h} dt/(1+t^2)$

Here it is again using \frac to write fractions a little nicer.
$f(x+h) - f(x) = \int_0^{x+h} \frac{dt}{1+t^2} - \int_0^x \frac{dt}{1+t^2} = \int_x^{x+h} \frac{dt}{1+t^2}$

How do I decide this lower and upper bound?

I will try latex in my next post :)

 

[Mark44]

How do I decide this lower and upper bound?

Do you know how to approximate an integral using a rectangle?

 

[Laurentiusson]

Do you know how to approximate an integral using a rectangle?

Well I have just started with it so I'm not really sure to be honest.

 

[Mark44]

So give it a try...

 

[LCKurtz]

Two theorems about integrals of continuous functions that should be in your text are:

If $m \le f(x) \le M$ on $[a,b]$ then $m(b-a)\le \int_a^b f(x)~dx \le M(b-a)$ and

and the Mean Value Theorem:

There exists $c$ in $[a,b]$ such that $\int_a^b f(x)~dx = f(c)(b-a)$.

Either of these should be useful to you.

 

[Laurentiusson]

So give it a try...

I guess it's this formula I should use:

f(x) = ∫_0^x dt/(1+t^2)

\int_x^{x+h} \frac{dt}{1+t^2}

Which one of the above should I insert in the formula?

 

[Laurentiusson]

I'(x) = (1/h) ∫_x^(x+h) dt/(1+t²)

I'(x) = (1/h) ∫_x^(x+h) dt/(1+t²) = (1/h) · (x+h-x) · 1/(1+c²)

where x ≤ c ≤ x+h.
c → x when h → 0

such as:

I'(x) = 1 · 1/(1+x²) = 1/(1+x²).

 

[Mark44]

I guess it's this formula I should use:

The hint in post #1 said to use the squeeze theorem, so you want to bound your integral by a value smaller than the integral and by a value larger than the integral. If you sketch a graph of y = 1/(1 + x²) you'll see that if x ≥ 0, this function is decreasing. Instead of taking a number c inside the interval [x, x + h], use the endpoints of the interval to get the heights of two rectangles.

f(x) = ∫_0^x dt/(1+t^2)

\int_x^{x+h} \frac{dt}{1+t^2}

Which one of the above should I insert in the formula?

Keep in mind what you're doing, which is using the definition of the derivative to find the derivative of the integral in post #1. Go back through this thread to remind yourself where $\int_x^{x+h} \frac{dt}{1+t^2}$ came from. Also note that you are missing a factor if 1/h that comes from the definition of the derivative.

Please use LaTeX in your work. I gave you some tips in post #16. You are already doing most of the work. All that is missing are two # characters at the front and two more # characters at the end.

 

[Laurentiusson]

$\int_x^{x+h} \frac{dt}{1+t^2}≈(x+h-x) f (c)≈(h) f (c)$

What should I do with h and my function c?

 

[Mark44]

$\int_x^{x+h} \frac{dt}{1+t^2}≈(x+h-x) f (c)≈(h) f (c)$

What should I do with h and my function c?

Please read my post, #24.

 

[Laurentiusson]

Please read my post, #24.

I have read it. I just don't understand what you mean and that'swhy I ask.

 

[Mark44]

In the squeeze theorem you set it up so that $f(x) \le g(x) \le h(x)$. If it turns out that $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then you can conclude that $\lim_{x \to a} g(x) = L$ as well.
$$(1/h) A \le (1/h) \int_x^{x + h} \frac{dt}{1 + t^2} \le (1/h) B$$

What can you put in for A and B? If you have a sketch of the graph of y = 1/(1 + t²) it should be fairly obvious how you can bound the integral.

 

[Laurentiusson]

It looks like my A is 1 and my B is 0 since the graph has a max at 1 and approaches 0 both to left and right.

 

[Mark44]

It looks like my A is 1 and my B is 0 since the graph has a max at 1 and approaches 0 both to left and right.

No. A and B aren't the max and min values of 1/(1 + t²). They are the max and min values of the integral.

 

[LCKurtz]

It looks like my A is 1 and my B is 0 since the graph has a max at 1 and approaches 0 both to left and right.

No. A and B aren't the max and min values of 1/(1 + t²). They are the max and min values of the integral.

And even so, you need to use the max and min of the function on the given interval of integration in your calculations.

 

[Laurentiusson]

I'm just more and more confused for every sentence you write. What am I actually suppoadd to do?

 

[LCKurtz]

After participating in and watching this thread, I don't think we can get you through this problem without working it for you, which is against forum policy. My suggestion is that it is time for you to have a face to face meeting with your teacher.

 

[Mark44]

After participating in and watching this thread, I don't think we can get you through this problem without working it for you, which is against forum policy. My suggestion is that it is time for you to have a face to face meeting with your teacher.

I agree. We can't go further than the hints we have given without working the problem for you, which as LCKurtz says, is against the policy in this forum.

I'm just more and more confused for every sentence you write. What am I actually suppoadd to do?

If you have a specific question about something we have written, ask it, but saying you are more and more confused doesn't give anything to go on.

 

[Laurentiusson]

I don't have a teacher to ask which is the reason to why I'm here.

I wonder how I can see the max and mins for 1/(1+t^2). I thought I just should look for extreme points when I plot it but I guess I'm wrong.

 

[Mark44]

I don't have a teacher to ask which is the reason to why I'm here.

I wonder how I can see the max and mins for 1/(1+t^2). I thought I just should look for extreme points when I plot it but I guess I'm wrong.

On the interval [x, x + h], what is the maximum value of 1/(1 + t²)? What is the minimum value of 1/(1 + t²)?