# Fundamental theorem of calculus

1. Mar 13, 2007

### kingwinner

I have a big test tomorrow and as I was reviewing, I encountered the following confusions. I hope that someone can help me out. I really appreciate for your help!

1) http://www.geocities.com/asdfasdf23135/cal0007.JPG
The answer is NO.
But when I differentiate both sides and using the fundamental theorem of calculus, I get f(x)=e^x, which is WRONG. Why does the theorem give me the wrong answer?

2) http://www.geocities.com/asdfasdf23135/cal0006.JPG
The example says that at x=-r, u=-pi/2 (but why not 3pi/2, or 7pi/2?)
Clearly, sin (-pi/2) = sin (3pi/2) = -1. Say, if I use 3pi/2 (instead of -pi/2), my final answer would be -pi(r^2)/2, which is not the correct answer...why is this happening? What is the problem?

3) For what values of c does the equation ln x = c(x^2) have exactly one solution? Justify fully.

Clearly c<0 is part of the answer.
For the case of c>0, there is one case where the curves y=ln x and y=c(x^2) intersect at only one point. But I need one more equation to solve for 2 variables. If the 2 curvesthey have a COMMON tangent at this point, then I got my second equation and I can solve for c. But how can I justify that at that point, they have a COMMON tangent (i.e. same slope)? I have no idea...

Can someone please explain?

2. Mar 13, 2007

### Dick

Put x=0 into your original equation for part 1). Do you see what's wrong? Sorry, no time to comment just now on the other parts.

Last edited: Mar 13, 2007
3. Mar 13, 2007

### AlephZero

1) The result is clearly false because the integral = 0 when x = 0, for any integrable function f. The right hand side = e^0 = 1.

You have assumed the false result is true and then differentiated. Starting from a false result, you can "prove" anything you like.

For example, assume x = 2x for all values of x: differentiate, and you "prove" that 1 = 2.

4. Mar 13, 2007

### JasonRox

2=1!

5. Mar 13, 2007

### Hurkyl

Staff Emeritus
It doesn't. The FTC says that

$$\frac{d}{dx} \left( \int_0^x f(t) \, dt \right) = f(x).$$

It does not claim that the only solution to

$$\frac{d}{dx} g(x) = f(x)$$

is

$$g(x) = \left( \int_0^x f(t) \, dt \right)$$.

6. Mar 13, 2007

### Hurkyl

Staff Emeritus
Carefully write down your argument. What does the FTC actually say? I think you made the classic mistake of reversing the problem. I think you proved

If $\int_0^x f(t) \, dt = e^x$, then $f(x) = e^x,$

but you mentally reversed it to the following incorrect statement:

If $f(x) = e^x$, then $\int_0^x f(t) \, dt = e^x.$

7. Mar 13, 2007

### kingwinner

I see what you mean, thanks!

1b) http://www.geocities.com/asdfasdf23135/cal0009.JPG

Differeniate both sides
=> f(x^2) * 2x = 4x * e^(2x^2)
Now can I cancel out the "x"? (x may possibly be 0, right?), and how can I proceed from there?

Help...

8. Mar 13, 2007

### theperthvan

There's no need for the factorial after the 1, because 1! = 1 anyway!(exclamation mark)

9. Mar 13, 2007

### kingwinner

Also, can anyone explain Q2 and Q3? Thanks!

10. Mar 13, 2007

### Hurkyl

Staff Emeritus
I wanted to add something. Maybe it was clear already, but I just want to make sure.

This reverse analysis is a good thing for analyzing a problem -- the things you derive tell you facts that must be true of any solution. In this case, it worked very well, because it shows that any solution must satisfy $f(x) = e^x$, so there is only one candidate solution. It just so happens that it isn't a solution, and thus the original equation has no solutions.

11. Mar 13, 2007

### Hurkyl

Staff Emeritus
If you have to worry about a special case, then work by cases! Consider separately the case where x is nonzero (and thus you may cancel) and the case where x is zero.

Well, you are seeking some sort of expression of the form "f(y) = ____", right? When you have an equation with an indeterminate value x, you can plug any expression you want in for x...

Last edited: Mar 13, 2007
12. Mar 13, 2007

### kingwinner

What happens to the case where x is zero? I can't cancel the "x", so I don't know how to proceed...

For the case where x isn't zero,
f(x^2) * 2x = 4x * e^(2x^2)
=>f(x^2) = 2 e^(2x^2)
Now is it valid to say that f(x)=2 e^(2x) ? Can I simply replace everywhere I see x^2 with x?

Will it run into the same kind of trouble as the quesiton
http://www.geocities.com/asdfasdf23135/cal0007.JPG
where f(x)=e^x is the wrong answer?

Is f(x) = 2 e^(2x) guaranteed to be the right answer? If not, why not?

Thanks for your help!

13. Mar 14, 2007

### Gib Z

You know instead of asking us, why don't you try to check your answer, see if your answer fulfils the question.

The first one in ur OP didn't, if you tried your answer you would have seen that it didn't work.

This time, you get f(x)=2e^(2x), so f(t)=2e^(2t), check if it satisifes it!

$$\int_0^x 2e^{2t} dt = e^{2x} - 1$$. Is that what your question wanted it to be?

Edit: Beaten to it..

14. Mar 14, 2007

### HallsofIvy

You certainlly CAN say :

Yes. It might make more sense for you to substitute y= x2 first:
f(y)= 2e2y and then, for course, f(x)= 2e2x.

Now, because the problem specifically said "continuous" you don't have to worry about x= 0 separately- you get that f(0)= 2 by continuity.

With the first problem, "Does there exist f(x) such that
$$\int_0^x f(t)dt= e^x$$?"
You were essentially arguing that IF f(x) exists such that
$$\int_0^x f(t)dt= e^x$$"
then, by the fundamental theorem of calculus, you must have
f(x)= (ex)'= ex. But then you put it into the integral and find that
$$\int_0^x e^t dt= e^x- 1$$
NOT ex. Therefore there is NO such function and the answer to the question is NO.

Here, you are arguing that if there exist a (continuous) function satisifying
$$\int_0^{x^2} f(t)dt= e^{2x^2}- 1[/itex] then by the fundamental theorem of calculus, we must have [tex]2xf(x^2)= 4xe^{2x^2}$$
and so
$$f(x)= 2e^{2x}$$
Now what happens if you plug that the integral? (And do you see why that "-1" is important?)

15. Mar 14, 2007

### AlephZero

Look at the equation. When x = 0, the both sides of the original equation are 0. So it doesn't hit the same problems as the first example.

Re the other questions:

2) You want a one-to-one correspondence between the points in the two intervals you are integrating over. If you add arbitrary multiples of 2 pi, then as u goes from u_min to u_max x will go from -r to +r several times, not just once.

3) Use the fact that the functions are continuous. As a sketch of a proof, let f(x) = ln(x) - cx^2 with c > 0. When x is very small or very large, f(x) < 0. There are 0 1 or 2 roots depending on whether the maximum value of f(x) is < 0, = 0, or > 0,

Last edited: Mar 14, 2007
16. Mar 14, 2007

### kingwinner

1b) In many problems of solving for something, I don't have to verify my answer; as soon as I didn't make any mistake, my final answer should definitely be correct. For example, solve x^2-2x+1=0 gives x=1 and I don't really have to sub. x=1 back in and check. (imagine in a more complicated question where the solution is ugly and you don't have a calculator, you still don't have to verify your answer provided that you have done everything correctly)

But why in this case I HAVE to always verify my answer?

17. Mar 14, 2007

### HallsofIvy

Actually, there are many algebraic problems in which if you do everything correctly, you can still get an incorrect answer. If you have a fraction problem and solve by multiplying the equation by the least common denominator you might get a wrong answer because your solution might make one of the denominators 0. Similarly, if you solve an equation involving square roots by squaring, you might get a solution that makes one of the square roots negative.

In this case, you can obviously apply the "fundamental theorem" only if such a function exists. The way the first problem was phrased was a clue to that.

18. Mar 14, 2007

### Hurkyl

Staff Emeritus
That only happens when your invidual steps truly are reversible. For example:

x + z = y + z​

is true if and only if

x = y​

To phrase it the way I did earlier, both of these statements are true:

If x + z = y + z, then x = y.​
If x = y, then x + z = y + z.​

You always have to "check" your answer when you try to solve a problem by simplifying it. But if you only use if-and-only-if simplifications, then you've already done the work you need to check it. (Because you really and truly can just reverse the simplifications)

19. Mar 14, 2007

### AlephZero

In the first question, what you did was to show that

If the function f(x) exists then it must be e^x

But you also need to show that the function does exist. One way to do that is to check your proposed e^x solution satisfies the conditions of the question - and it doesn't. Therefore, there is no possible solution.