Fundamental theorem of calculus (something isn't right)

[njama]

The 2nd part of fundamental theorem of calculus says:

If f is continuous on an interval I, then f has an anti derivative on I. In particular, if a is any number in I, then the function F defined by
F(x)=\int_{a}^{x}f(t)dt

is an antiderivative of f on I; that is F'(x) = f(x) for each x in I.

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:

If f is continuous on [a,b] and F is any anti derivative of f on [a,b], then
\int_{a}^{b}f(x)dx=F(b)-F(a)

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.

Thanks in advance.

 

[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.

 

[njama]

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.

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:

<br /> F(x)=\int_{1}^{x}\frac{dt}{t}<br />

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?

 

[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.

 

[njama]

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.

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:

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 not because the derivative of it, 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 f(x)=|2x-3| is defined at all real numbers.
So that F&#039;(x) = f(x) except for x=3/2.

Am I right?

Thank you.

 

[espen180]

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.

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).

 

[njama]

Thank you, I understand now.

In that case:
<br /> F(x)=\frac{|2x-3|x(x-3)}{2x-3}<br />
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 (-∞,∞)?

 

[espen180]

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

 

[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?

 

[espen180]

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

 

[njama]

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

<br /> F(x)=\frac{|2x-3|x(x-3)}{2x-3}<br />

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?

 

[espen180]

Yes, that looks right.

 

[njama]

Yes, that looks right.

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 textbook 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.

 

[statdad]

Well your first definition of the theorem is quite 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)
.

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

If x < 3/2 this simplifies to

<br /> F(x) = -x(x-3)<br />

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).

 

[statdad]

The function

<br /> F(x) = \int_1^x \frac 1 t \, dt<br />

is not defined on (-\infty, 0) since, if x &lt; 0, the integral

<br /> \int_1^x \frac 1 t \, dt<br />

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

 

[statdad]

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.

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

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

is

<br /> F&#039;(x) = |2x-3|, \quad x \ne 3/2<br />

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

<br /> g(x) = |2x - 3|<br />

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

<br /> g(x) = \begin{cases}<br /> -(2x-3) &amp;\text{ if } x &lt; 3/2 \\<br /> \hphantom{-}(2x-3) &amp;\text{ if } x \ge 3/2<br /> \end{cases}<br />

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.

 

[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:

If f is continuous on interval I then it has anti-derivative on I

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

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}<br /> x(x-3), x &gt; \frac{3}{2}\\ <br /> \frac{-9}{4}, x=3/2 \\<br /> -x(x-3)-9/2,x&lt;\frac{3}{2}<br /> \end{matrix}\right.

 

[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 I am very sure that not only is 2t-3 integrable from 0 to 2,
its antiderivative t^2 - 3t + c is also

 

[elibj123]

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

If x < 3/2 this simplifies to

<br /> F(x) = -x(x-3)<br />

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).

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

<br /> F(x)=\frac{|2x-3|x(x-3)}{2x-3}<br />

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.

 

[statdad]

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

Apparently I understand this issue better than you - so let's stop this back and forth.

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

<br /> F(x)=\frac{|2x-3|x(x-3)}{2x-3}<br />

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.

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.

 

[statdad]

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.


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

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.

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.

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}<br /> x(x-3), x &gt; \frac{3}{2}\\ <br /> \frac{-9}{4}, x=3/2 \\<br /> -x(x-3)-9/2,x&lt;\frac{3}{2}<br /> \end{matrix}\right.

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

<br /> F&#039;(x) = \begin{cases}<br /> \hphantom{-(}2x - 3\hphantom{)} \quad x \ge 3/2 \\<br /> -(2x-3) \quad x &lt; 3/2<br /> \end{cases}<br />

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.

 

[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.

 

[elibj123]

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.

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

<br /> F(x)=\frac{|2x-3|x(x-3)}{2x-3}<br />

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

<br /> F(x)=x(x-3); x&lt;3/2<br />
<br /> F(x)=-x(x-3)+4.5; x&gt;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.

 

[statdad]

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.

 

[njama]

Thanks for the replies.

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.

@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.

 

[HallsofIvy]

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

That is, of course, not true as long as we are talking about real numbers.

f(x)= \frac{1}{x^2+ 1}

has an "x in the denominator" but is continuous for all real numbers.

 

[snipez90]

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.

All right fine I'll do a Riemann integration proof.

Claim:
If f is integrable on [a,b] and we define F to be F(x) = \int_{a}^{x}f, then F is continuous on [a,b].
Proof:
If f is Riemann integrable, then f is bounded, so there exists M > 0 such that f(t) \leq M for all t in [a,b]. Suppose a\leq x &lt; y \leq b.
Then
|F(y) - F(x)| = \int_{x}^{y}f \leq M(y-x),
where the last inequality follows easily from the theory of Riemann/Darboux sums, or any old calculus book (it's also intuitively obvious).
Let \varepsilon &gt; 0 be given and choose \delta = \frac{\varepsilon}{M}. If |y-x| &lt; \delta, then
|F(y)-F(x)| &lt; \varepsilon.
This shows that F is not only continuous, but actually uniformly continuous (it's actually Lipschitz continuous, but who cares).
Note there is no mention about the continuity of f. For f to be Riemann integrable, it is necessary and sufficient for it to be discontinuous on a set of measure zero.

 

[njama]

The condition for continuity of F is:

\lim_{x \rightarrow x_0}F(x)=F(x_0)

if given any number \varepsilon &gt; 0 [/tex] we can find a number \delta &amp;gt;0[/tex] such that |F(x)-F(x_0)| &amp;amp;lt; \varepsilon if 0 &amp;amp;lt; |x-x_0| &amp;amp;lt; \delta.&lt;br /&gt; &lt;br /&gt; Because f(x) is defined on interval [a,b] and it is continuous on the interval then by the extreme-value theorem f(x) have minimum and maximum.&lt;br /&gt; m \leq f(x) \leq M&lt;br /&gt; &lt;br /&gt; m \leq \frac{1}{b-a} \int_{a}^{b}f(x)dx \leq M&lt;br /&gt; &lt;br /&gt; |F(b)-F(a)|= \int_{a}^{b}f(x)dx \leq M (b-a)&lt;br /&gt; &lt;br /&gt; if&lt;br /&gt; &lt;br /&gt; 0 &amp;amp;lt; |b-a| &amp;amp;lt; \delta&lt;br /&gt; &lt;br /&gt; and&lt;br /&gt; &lt;br /&gt; 0 &amp;amp;lt; M|b-a| &amp;amp;lt; M * \delta&lt;br /&gt; &lt;br /&gt; if M&amp;gt;0&lt;br /&gt; &lt;br /&gt; so that&lt;br /&gt; &lt;br /&gt; |F(b)-F(a)|= \int_{a}^{b}f(x)dx \leq M (b-a) &amp;amp;lt; M*\delta = \varepsilon&lt;br /&gt; &lt;br /&gt; Yes, the proof is correct. &lt;br /&gt; &lt;br /&gt; BUT, on the start you defined &lt;br /&gt; &amp;lt;br /&amp;gt; F(x) = \int_{a}^{x}f,&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; What that statement means ?&lt;br /&gt; &lt;br /&gt; Its just different notation of F&amp;#039;(x)=f(x). So if f(x) is continuous on interval I this implies F&amp;#039;(x) is continuous on I and this implies F(x) is continuous on I. Clearly at the end the continuity of f(x) will imply the continuity of F(x). Do you agree?&lt;br /&gt; &lt;br /&gt; That is my simple proof.

 

[jgens]

You seem to be confused about the definition of the integral: F(x) = \int_a^xf is not just different notation for F&#039;(x) = f(x).

 

[njama]

Ok. I used wrong words. I meant to say that integration is opposite process of derivation and vice versa.

By stating that
<br /> F(x) = \int_a^xf<br /> you also state that F'(x)=f(x)

 

[jgens]

Nope. Anti-differentiation is the reverse process of differentiation but integration is another thing entirely. The fundamental theorem of calculus links integration and anti-differentiation but they are by no means the same thing.

 

[njama]

Nope. Anti-differentiation is the reverse process of differentiation but integration is another thing entirely. The fundamental theorem of calculus links integration and anti-differentiation but they are by no means the same thing.

Sorry for going out of the boundaries of this thread, but could you possibly tell me what's the difference?

 

[jgens]

Sorry for going out of the boundaries of this thread, but could you possibly tell me what's the difference?

Alrighty, in order to avoid any confusion with the integral sign, I'm going to introduce notation that is definitely nonstandard, so don't use it anywhere else. Let D be the differential operator so that D(f) = f&#039; and let D^{-1} denote an anti-differentiation operator where D^{-1}(D(f)) = D^{-1}(f&#039;) = f. The integral sign \int will be used solely for the purpose of integration in this example.

With the premliminary information ouf of the way, I suppose that I'll start with anti-differentiation. As the name suggests, this process is simply the reverse of differentiation. Hence, if D(f) = f&#039; then D^{-1}(f&#039;) = f and if D^{-1}(f) = F then D(F) = f. Hopefully you can understand this concept fairly easily.

Now, integration on the other hand, is an entirely different process. Suppose that we have the function f which is defined on the closed interval [a,x]. If P = \{t_0, \dots, t_n\} is a partition of [a,x] such that a = t_0 &lt; t_1 &lt; \dots &lt; t_{n-1} &lt; t_n = x then we define the Riemann sum of f for the partition P by R(f,P) = \sum_{i=1}^{n}f(x_i)(t_i - t_{i-1}) where t_{i-1} &lt; x_i &lt; t_i. With this information, the integral of f is given by

\lim_{||P|| \to 0}R(f,P) = \lim_{n \to \infty}R(f,P) = \int_a^xf

Note that D^{-1} and \int have two different definitions. Alright, we're almost done now. What the Fundamental Theorem of Calculus states is that if f is continuous and we define the function F by F = \int_a^xf, then F is differentiable and in particular D(F) = f. Hopefully someone else here can give you a better explanation of everything, but this is the best that I can do with the limited time that I have. Good luck!

 

[HallsofIvy]

A problem with that idea is that "D^{-1}" is not well defined. Given f(x), there may exist and infinite number of functions, F such that DF= f.

 

[jgens]

Halls, I realize that it's poor notation (hence my note about it early on), but it eliminates the need for the integral sign when talking about anti-derivatives. I figured that this might be useful for a student who was struggling with the difference between anti-differentiation and integration. I suppose that I could have given my explanation without introducing some sort of notation for anti-differentiation though. If you have the time to write up an explanation for the OP I'm sure that he/she will appreciate it since you'll do a far better job explaining it then I'll ever do.

 

[njama]

Thanks for the explanation jgens. I partially understand what are you saying to me, but I can't really see the big difference. Suppose [-1,x] is interval on which I need to find the area on. So A(x) would be the area. But it can be seen that \int_{f(x)dx}=A(x) or A'(x)=f(x). I also checked wikipedia.

Here is what I found:

In calculus, an antiderivative, primitive or indefinite integral[1] of a function f is a function F whose derivative is equal to f, i.e., F ′ = f. The process of solving for antiderivatives is antidifferentiation (or indefinite integration)

 

[jgens]

If you want a simple illustration of how they're different, consider the function f defind by f(x) = 0 if x = 1 and f(x) = 1 otherwise. Now suppose that I want to evaluate \int_0^xf. It's a trivial exercise to show that \int_0^xf = x. Now if integration were the inverse process of differentiation we would have that (x)&#039; = 1 = f(x) which clearly isn't true. Does this make sense?

 

[njama]

If x=1:
<br /> \int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C<br />
If x\neq 1
\int_0^x1dt=(t+C_1)|_0^x = x-0=x
It still valid:
F(x)=\left\{\begin{matrix}<br /> C, x=1\\ <br /> x, x \neq 1<br /> \end{matrix}\right.

f(x)=F&#039;(x)=\left\{\begin{matrix}<br /> 0, x=1\\ <br /> 1, x \neq 1<br /> \end{matrix}\right.

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

 

[elibj123]

If x=1:
<br /> \int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C<br />
If x\neq 1
\int_0^x1dt=(t+C_1)|_0^x = x-0=x
It still valid:
F(x)=\left\{\begin{matrix}<br /> C, x=1\\ <br /> x, x \neq 1<br /> \end{matrix}\right.

f(x)=F&#039;(x)=\left\{\begin{matrix}<br /> 0, x=1\\ <br /> 1, x \neq 1<br /> \end{matrix}\right.

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

You cannot just swap f(x) for zero, because the integral "doesn't ask" the function what's its value on a certain point, but what's its value on an interval. Therefore the integral goes through "many" points in which f(x)=1, bumps into a zero, and then goes on. (This is a point where I recommend to read again about integrals, and exercise them, because what you did in the first lines "if x=1 blahblah" show you don't completley understand how to work with them and what do they mean)
But we know that changing a function in a finite number of points, doesn't change its integral, so as far as the integral "knows" he's actually integrating only f(x)=1.

Anti-differentiation, if studied as an operator, must consider any information in f(x), while integration loses this little things that make up f(x).

 

[jgens]

If x=1:
<br /> \int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C<br />
If x\neq 1
\int_0^x1dt=(t+C_1)|_0^x = x-0=x
It still valid:
F(x)=\left\{\begin{matrix}<br /> C, x=1\\ <br /> x, x \neq 1<br /> \end{matrix}\right.

f(x)=F&#039;(x)=\left\{\begin{matrix}<br /> 0, x=1\\ <br /> 1, x \neq 1<br /> \end{matrix}\right.

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

No, no, no! This isn't right in the slightest. You can't just integrate using the FTC if f isn't continuous. You need to go back using the definition of the integral like the one that I provided and not what you mistakenly keep taking as the definition of the integral.

Anyway, once you actually do the exercise correctly, you will find that \int_0^xf = x; hence, even though f isn't continuous, this integral is. Now, if integration and anti-differentiation were the same thing, differentiating \int_0^xf would show that f(x) = 1 which is clearly false. Therefore, anti-differentiation and integration are not the exact same thing.

 

[njama]

OK. You defined the function like this:

And you told me to find:

<br /> \int_0^bf<br />

Now. If the function is continuous on interval [0,b] then it is integrable on [0,b].

But if function is bounded and have finitely many discontinuity on [0,b] it is still integrable.

So, we can find
<br /> \int_0^bf<br />

Lets consider any partition on [0,b]. Then either x_{k}^{*}=0 or it does not.
If not then
\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k} = \sum_{k=1}^{n}\Delta x_{k} = b
else
\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k} = -\Delta x_{k} + \sum_{k=1}^{n}\Delta x_{k} = b - \Delta x_{k}

which means that the difference between the Riemann sum and b is at most \Delta x_{k}. BUT, since \Delta x_{k} approaches zero as max \Delta x_{k} \rightarrow 0 it follows that:

\int_{0}^{b}f(x)dx = b

But the Fundamental Theorem of Calculus Part 2 states that "If f is continious on an interval [0,b] then f has an antiderivative on [0,b]. In particular if a is any number in [0,b] then the function F defined by:
F(x)=\int_{a}^{x}f(t)dt
is an antiderivative of f on [0,b]; that is F'(x)=f(x) for each x in [0,b], or in an alternative notation:
\frac{d}{dx}\left [ \int_{a}^{x}f(t)dt \right ] = f(x)

So f must be continious so that we can conclude that f has an antiderivative. If f is not continious anything can follow.

 

[jgens]

So f must be continious so that we can conclude that f has an antiderivative.

Except for the fact that f clearly isn't continuous so the FTC isn't applicable here. The point of this exercise was to illustrate that if F is a function defined by F = \int_0^xf, then F&#039; is not necessarily equal to f as you asserted in post number 30 of this thread.