# Fundamental theorem of calculus (something isn't right)

1. Feb 12, 2010

### njama

The 2nd part of fundamental theorem of calculus says:
Over what open interval does the formula

$$F(x)=\int_{1}^{x}\frac{dt}{t}$$

represent antiderivative of f(x)=1/x ?

By looking at the theorem I would say that f(x) is continuous only for $$x \neq 0$$
So I would say that F(x) is defined on $$(-\infty, 0) U (0, +\infty)$$ that is F'(x)=f(x) for all numbers $$x \neq 0$$.

But there is problem. F(x) actually equals Log(x) , because (Log(x))' = 1/x.

Why this theorem does not work?

Now there is problem with the fundamental theorem of calculus part 1:

Here is the counterexample:

$$\int_{0}^{2}|2x-3|dx=\frac{|2x-3|x(x-3)}{2x-3}|_{0}^{2}$$

But again the area is not -2 it is actually 5/2.

One more thing:

$$(\frac{|2x-3|x(x-3)}{2x-3})' = |2x-3|$$ where $x \neq 3/2$ because F(x) is not defined at 3/2

Which is not same as |2x-3| which is defined everywhere.

2. Feb 12, 2010

### elibj123

Well your first definition of the theorem is quite wrong.

A more accurate form will be:

Let f(x) be integrable on some interval I, and let a be a real number in I. Then whenever
the function

$$F(x)=\int^{x}_{a}f(s)ds$$

Is continuous at a point x0, its derivative at this point will be equal to the original function: $$F'(x_{0})=f(x_{0})$$

Otherwise, if F is not continuous or defined at the point, the theorem fails.

As for the second part, notice that the formula requires a function F which is continuous in [a,b], so it will be the real anti-derivative. Otherwise, again, the theorem fails.

Of course your F is discontinuous at x=3/2: F(3/2-)=-3/2(3/2-3). F(3/2+)=3/2(3/2-3)
And therefore your result should be wrong.
But at any interval [0,3/2), (3/2,2] you may choose an anti-derivative up to a constant.
Therefore there exist such constant that at x=3/2 the functions will be continuous (while not defined, it will have a removable discontinuity, which happens to work)

Perhaps this will help you understand what was bothering you. I didn't quite catch it at the first part of the post.

3. Feb 12, 2010

### njama

Thanks a lot for the reply.

Now it make my things clearer by stating that F(x) must be continious so that it would be anti-derivative.

Now I know that $$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$ is not the actual anti derivative of $$|2x-3|$$ and it is because the derivative of is not defined at 3/2, so that when I find F'(x) it would be equal to |2x-3| for $$x\neq 3/2$$

and |2x-3| is defined at all real numbers. Am I right?

By writing the definite integral as:

$$\int_{0}^{2}|2x-3|dx=\int_{0}^{3/2}(2x-3) + \int_{3/2}^{2}-(2x-3)$$

I somehow "skip" the discontinuity at 3/2.

For the first part. I guess the theorem lacks of saying that F(x) must be defined on the interval I.

For example. If I use the first theorem I would say that:

$$F(x)=\int_{1}^{x}\frac{dt}{t}$$

F(x) is defined for $$x\neq 0$$ because f(x) is continuous for all reals except x=0, and the theorem says that if f(x) is continuous on interval I then also F(x) is continuous on interval I.

But this is not true.

Actually f(x)=1/x is continuous for all reals except x=0.

BUT, F(x) = Log(x) and it is not continuous on all negative real numbers.

I know that also F(x) could be Log|x| but it also could be Log(x).

What is the problem? Is it with the book that I am using for learning?

4. Feb 12, 2010

### espen180

The problem is that you are not integrating 1/x correctly.

The integral of 1/x is ln|x|, not ln x. It just happens that for positive x, ln|x|=ln x.

5. Feb 12, 2010

### njama

But it can be ln(x). Why? Because (ln(x))' = 1/x. Because integration is anti-differentiation:

$$\int 1/x = ln(x)$$

Am I right?

Could you please possibly confirm my statement:

Thank you.

6. Feb 12, 2010

### espen180

Your argument is flawed. You start with a function which by is defined on (0,∞). When you differentiate this fuction, it will only be valid on (0,∞), even if it is defined on (-∞,∞). Thus, you cannot expect the integral of 1/x to be ln(x) outside (0,∞).

To account for the fact that 1/x is defined on (-∞,∞), its integral is ln|x|, not ln(x).

7. Feb 12, 2010

### njama

Thank you, I understand now.

In that case:
$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$
is defined on real numbers except x=3/2

So on F'(x)=|2x-3|=f(x) we cannot expect f(x) to be defined on x=3/2 since F'(3/2) is not defined because of the discontinuity of F(x) on x=3/2.

But what we need to make to let |2x-3| be defined on (-∞,∞)?

8. Feb 12, 2010

### espen180

Any fuction with x'es in the denominator will have a discontinuity and thus can't be defined for all x.

9. Feb 12, 2010

### njama

Does that mean that we need to set notice that the function f(x)=|2x-3| is valid on all real numbers except x=3/2, before stating the integral:

$$\int{|2x-3|dx}$$

so that F(x) would also not be defined on x=3/2?

10. Feb 12, 2010

### espen180

That fuction itself has no problem with being defined for all real x. why should it? |0| is just 0.

11. Feb 12, 2010

### njama

$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$

F(x) is not defined on x=3/2 so that F'(x) would be not defined on x=3/2 so that F'(x)=f(x) would be only valid for $$x \neq 3/2$$, right?

12. Feb 12, 2010

### espen180

Yes, that looks right.

13. Feb 12, 2010

### njama

Ok, that means for

$$f(x)=|2x-3|$$ is valid only for $$x \neq 3/2$$ (even it is continuous on infinity), because of the discontinuity of its anti-derivative:

$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$

And that is the problem with the theorem which says:
"If f is continuous on an interval I, then f has an anti derivative on I."

In this case f is continuous on infinity, but the anti derivative is not defined on infinity.

elibj123 gave the correct theorem, but how is possible that book for Calculus from Irl Bivens, Steven Davis and Howard Anton made such a mistake?

Also in the solutions of the text book they say that:

$$F(x)=\int_{1}^{x}\frac{dt}{t}$$

is defined on open interval (0, +∞) when it is defined on (-∞,0) U (0,+∞)?

It is the same for:

$$F(x)=\int_{1}^{x}\frac{dt}{t^2-9}$$

except for the interval (3,+∞) whether it should be (-∞,3) U (3,+∞).

They wrote the theorem and disregard it:

"If f is continuous on an interval I, then f has an anti derivative on I."

even it is incorrect.

14. Feb 12, 2010

No - the function used does not have a removable discontinuity at x = 3/2. the function is
$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$

If x < 3/2 this simplifies to

$$F(x) = -x(x-3)$$

For x > 3/2 it simplifies to $$F(x) = x(x-3)$$. You cannot provide any value at x = 3/2 to turn this into a continuous function (try graphing it to see this).

15. Feb 12, 2010

The function

$$F(x) = \int_1^x \frac 1 t \, dt$$

is not defined on $$(-\infty, 0)$$ since, if $$x < 0$$, the integral

$$\int_1^x \frac 1 t \, dt$$

does not exist (it doesn't converge). $$F$$ IS continuous on $$(0,\infty)$$

16. Feb 12, 2010

I have no idea what you mean by defined on infinity''. Your problem comes from the way you are dealing with $$|2x-3|$$.

Think this way. You are saying that the derivative of

$$F(x) = \frac{|2x-3|x(x-3)}{2x-3}, \quad x \ne 3/2$$

is

$$F'(x) = |2x-3|, \quad x \ne 3/2$$

and that is true - but you have to stick with the condition $$x \ne 3/2$$. In this case, you can say this: since |2x-3| is defined and continuous on the interval $$(-\infty, 3/2) \cup (3/2, \infty)$$, it has a derivative there (one that works is your function F).

What you do in the second part is to consider the function

$$g(x) = |2x - 3|$$

as a starting point. This function is continuous for all real numbers, so it has an antiderivative. This is not the contradiction you claim: in the case above the absolute value function wasn't defined at 3/2; its antiderivative was F. The current function is defined at 3/2, so its antiderivative is defined at 3/2 and so the antiderivative IS NOT THE FUNCTION F. Here

$$g(x) = \begin{cases} -(2x-3) &\text{ if } x < 3/2 \\ \hphantom{-}(2x-3) &\text{ if } x \ge 3/2 \end{cases}$$

An antiderivative can be formed from this.

In summary: in the first situation you arrived at |2x-3| by considering it to be derivative function not defined at 3/2, so in this context it is not appropriate to treat |2x -3| as though it is defined at 3/2.

In the second part you start with |2x-3| as its own function, so it is defined and continuous everywhere; that means it has an antiderivative everywhere, but that antiderivative cannot be F. There is no contradiction with the theorem.

17. Feb 13, 2010

### njama

Thanks a lot for the posts.

First you're right by saying that the anti-derivative is not F(x).

But the theorem says:

F(x) is an anti-derivative of f on I, that is F'(x) = f(x) for each x in I.

Also there is theorem which says:

This is not true.
$$F(x)=\int_{1}^{x}|2t-3|dt$$

f(t)=|2t-3| is continuous on [0,2] but F(x) is not continuous on [0,2] it has gap in x=3/2.

I am just following the theorem. That's the problem with the theorem.

The theorem must state that F(x) must be continuous so that F'(x)=f(x) will work for all x in the interval. Do you agree?

Or maybe I found such F(x) which is continuous on any interval [a,b]

$$F(x)=\left\{\begin{matrix} x(x-3), x > \frac{3}{2}\\ \frac{-9}{4}, x=3/2 \\ -x(x-3)-9/2,x<\frac{3}{2} \end{matrix}\right.$$

Last edited: Feb 13, 2010
18. Feb 13, 2010

### raymo39

if f(x) is continuous from x=a to x=b, then f(x) is integrable from a to b.
IF F(x) (the antiderivative of f(x)) is not continuous from a to b, then it will not be integrable from a to b
but im very sure that not only is 2t-3 integrable from 0 to 2,
its antiderivative t^2 - 3t + c is also

19. Feb 13, 2010

### elibj123

You missread my post and perhaps didn't understand the subject yourself.

The antiderivative of a function is not unique, a function has infinitely many anti-derivatives differing by a constant. That constant doesn't have to be "constant" on the entire interval.

For example

F(x)=2x+3 for x>0
F(x)=2x+100 for x<0

Is still an antiderivative of f(x)=2, but it's not of the form F(x)=2x+c (c changes between the intervals).

What I said is that while

$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$

is discontinuous, you can choose such c's, that the function F(x) (while not defined with a single formula) will be now continuous on the entire interval (neglecting the minor removable discontinuity at x=3/2) and then the definite integral identity applies.

20. Feb 13, 2010

Apparently I understand this issue better than you - so let's stop this back and forth.
No, you cannot make F(x) continuous by defining its value at 3/2. Perhaps you should graph it to see this. This is not a case where there is a hole in the graph but the two one-sided limits agree: the one-sided limits at 3/2 are different.

Last edited: Feb 13, 2010
21. Feb 13, 2010

No, you are not following the theorem - you are repeating the same error I mentioned earlier. When you refer to F as an antiderivative of |2x -3|, you must accept the fact that |2x-3| is not defined at 3/2, since F is not defined their. The use of |2x-3| is determined from the way in which it is obtained, and because if its origin in this instance, it IS NOT continuous for all real numbers, since 3/2 is not in its domain. There is no contradiction.

This function is continuous at 3/2, but notice that it is not the same function F you started all of this with. It is continuous, and differentiable at 3/2 (the two one-sided derivatives agree there), and its derivative is

$$F'(x) = \begin{cases} \hphantom{-(}2x - 3\hphantom{)} \quad x \ge 3/2 \\ -(2x-3) \quad x < 3/2 \end{cases}$$

which is a long'' way to write $$F(x) = |2x-3|$$. Since this function F is continuous everywhere, and differentiable, the theorem holds and there is no contradiction.

22. Feb 13, 2010

### snipez90

Hmmm the theorems are stated incorrectly many times in this thread, and this might be a source of confusion.

First of all, if f is merely integrable on [a,b] and we define F to be $F(x) = \int_{a}^{x}f,$ then F is continuous on [a,b].

The "theorem" suggested in post 2 was simply wrong. F is always continuous provided that f is integrable. This is not the fundamental theorem of calculus.

The fundamental theorem tells us that if f is continuous (not merely integrable), then F is differentiable and F'(c) = f(c) for c in (a,b). Again, the continuity of f here is essential. We already know F is continuous, but differentiability of F depends on continuity of f.

23. Feb 13, 2010

### elibj123

I said nothing about defining its value at 3/2.
I was talking about adding different constants to it in the different intervals, so the one-sided limits will agree.

Of course it will not be of the form you wrote, it will have another definition, but it will remain the anti-derivative of f(x).

Oh, I've took your advise, and let me show that to you graphically (attached).

Now the red graph (it's supposed to have a branch on the left too)
If I'd simply defined it with

$$F(x)=\frac{|2x-3|x(x-3)}{2x-3}$$

But the blue graph is what would happen if I'd played with constants a little bit and would define like that:

$$F(x)=x(x-3); x<3/2$$
$$F(x)=-x(x-3)+4.5; x>3/2$$

You see, F is still an anti-derivative, yet now it's also continuous.

I still believe you misread my original post, no intension to insult, I'm sure we both understand the subject very well.

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24. Feb 13, 2010

I acknowledged the second form for the antiderivative in my post @21. I did completely miss the point of your initial explanation of "fixing" the antiderivative with constants - I apology.

25. Feb 14, 2010

### njama

Thanks for the replies.
@snipez90 I don't agree with you.

If a function is continuous on [a,b] than it is integrable on [a,b] or F'(x)=f(x) for x in [a,b].

The condition of f to be continuous is necessary factor so that F'(x) to be differentiable AND this is the main factor that F(x) is continuous.

We do not know that F(x) is continuous until we know the fact that it is differentiable.

And does the reverse works?

And what if we suppose that F'(x)=f(x) on interval I. Then we can say for sure that F is differentiable and continuous on I. Also we can say that f is continuous on the interval I.