# Show (dt)/(1+t^2)

1. Mar 25, 2015

### Laurentiusson

• Moved from a technical section, so missing the template
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?

2. Mar 25, 2015

### LCKurtz

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.

3. Mar 25, 2015

### Laurentiusson

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

4. Mar 25, 2015

### 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.

5. Mar 25, 2015

### vela

Staff Emeritus
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.

6. Mar 25, 2015

### Laurentiusson

How to do this?

7. Mar 25, 2015

### LCKurtz

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.

8. Mar 25, 2015

### Laurentiusson

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

9. Mar 25, 2015

### LCKurtz

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

10. Mar 25, 2015

### Laurentiusson

That derivate would be -((2t)/(1+t2)2

11. Mar 25, 2015

### 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.

12. Mar 25, 2015

### Laurentiusson

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?

13. Mar 25, 2015

### LCKurtz

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.

14. Mar 25, 2015

### Staff: Mentor

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".

15. Mar 25, 2015

### Laurentiusson

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?

16. Mar 25, 2015

### Staff: Mentor

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.
$f(x+h) - f(x) = \int_0^{x+h} \frac{dt}{1+t^2} - \int_0^x \frac{dt}{1+t^2} = \int_x^{x+h} \frac{dt}{1+t^2}$

17. Mar 25, 2015

### Laurentiusson

How do I decide this lower and upper bound?

I will try latex in my next post :)

18. Mar 25, 2015

### Staff: Mentor

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

19. Mar 25, 2015

### Laurentiusson

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

20. Mar 25, 2015

### Staff: Mentor

So give it a try...

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