# Homework Help: Interesting piece-wise problem

1. May 6, 2010

### mathgal

I hate a piece-wise function

f(x)= [ 1 if 0 </eq x </1
[ 0 if 1 < x < 2
[ -2x+4 if x >/eq 2

If I let F(x)= integral (from 0->x) f(t) dt, how can i find a formula for F(x) without using integrals? I think it may have something to do with the 2nd definition of the Fundamental Theorem of Calculus, but I'm not sure.

2. May 6, 2010

### Staff: Mentor

Have you graphed this function? After you do this, you should be able to come up with a formula for F(x), using nothing more sophisticated than geometry.

3. May 6, 2010

### mathgal

Should I be using limits? I actually did graph it. I know what the F(x) should be, I'm just not sure how to rigorously show it using a theorem, formula, etc

4. May 6, 2010

### Staff: Mentor

No, you don't need limits. This is mostly a geometry problem. Your formula for F(x) will be defined in a piecewise fashion.
To get you started,
F(x) = x, 0 <= x < 1

You will need two more definitions, one for each of the two additiona intervals. The formula for F(x) for 1 <= x < 2 is very easy.

5. May 6, 2010

### mathgal

Is this correct? I'm worried the last one should be more complicated?

F(x) = x, 0 <= x <= 1
= 0, 1 < x < 2
= -x^2+4x, x >= 2.

6. May 6, 2010

### mathgal

Well it should actually be F(x)=-x^2+4x-4 for x>= 2 right?

7. May 6, 2010

### The Chaz

Just one??? I hate ALL piece-wise functions!

8. May 6, 2010

### mathgal

Also, is there a way for me a find a formula for F'? How do I first find where F is differentiable?

9. May 6, 2010

### Staff: Mentor

The second and third parts are wrong. Remember that F(x) is defined as
$$\int_0^x f(t) dt$$

So, in particular, F(1.5) $\neq$ 0, as you have, and F(2) $\neq$ 4, as you have. F(2) should actually be equal to 1.

10. May 6, 2010

### Staff: Mentor

I divide functions into three categories:
• the ones I hate
• the ones I love
• the ones that I feel neutral about

11. May 6, 2010

### mathgal

so should F(x)=x also if 1<x<2?
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 0

is that correct?

12. May 6, 2010

### mathgal

Actually shouldn't F(x)= 1, 0<=x<=1, F(x)=1, 1<x<2, and F(x)=-x^2+4x-3, x>=2

13. May 6, 2010

### Staff: Mentor

Edit: This refers to post 11.
No, that's not it either. For example, what is F(1.5)? It's not 1.5.
I'm going to translate part of the line above into actual mathematics.
$$F(2) = \int_0^1 1~dx + \int_1^2 0~dx$$
The value of the first integral is not x, and x + 0 is not 0.

14. May 6, 2010

### Staff: Mentor

Well, you're close. The first part should be F(x) = x, 0 <= x <= 1.

Last edited: May 6, 2010
15. May 6, 2010

### mathgal

But we're evaluating it from (0,1)... So once we integrate and get x|, then we have 1-0=1. This isn't correct?

16. May 6, 2010

### mathgal

Oh and in Post #11, I meant to say:
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

17. May 6, 2010

### mathgal

Oh and in Post #11, I meant to say:
F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

18. May 6, 2010

### Staff: Mentor

A lot of this is gibberish. F(x) is not the integral from 1 to 2 and it's not the integral from 0 to 2, and it's not x, which is not 1.

F(x) needs to be defined in piecewise fashion.

19. May 6, 2010

### mathgal

okay well thanks, anyways

20. May 6, 2010

### Staff: Mentor

I'm not trying to discourage you (although I might be succeeding), but I am trying to get you to write things that make sense.

21. May 6, 2010

### mathgal

F(x)= x if 0 <= x <= 1.

Then, for 1< x <2, we're taking the integral of 0, so I should just get a constant?
I just don't know how I am supposed to write that part of the third part whre x>=2

22. May 6, 2010

### Staff: Mentor

$$F(x) = \int_0^x f(t) dt$$
If 0 <= x <= 1, F(x) = x
If 1 < x <= 2, F(x) = 1 + 0 = 1. The 1 comes from the rectangle above the interval [0, 1].
For the 2nd and 3rd intervals you have to add the contribution(s) from the previous interval(s).

It's really important to have a good graph of the function F, showing the three parts. The first part is a line with slope 1 between (0, 0) and (1, 1). the second part is a horizontal line between (1, 1) and (2, 1). The third part is the right half of a parabola that opens downward and whose vertex is at (2, 1). At the point where the parabola crosses the x-axis, F of that x is zero. From that point onward, F(x) < 0.

23. May 6, 2010

### mathgal

I think I actually understood that better than I thought and was just getting confused by my own writing. Thank you so much. I know you've already helped alot, but do you have any ideas about how I would then find a formula for F'(x) wherever F is differentiable? My test is tomorrow, that is the only reason I'm trying to learn this stuff now.

24. May 6, 2010

### Staff: Mentor

The graph of F is continuous for x >= 0. The only points where it is not differentiable are at x = 1 and x = 2. The formula for F'(x) will have three parts.

For 0 < x < 1, F(x) = x, so F'(x) = 1. Do the same for the other two parts, but keep in mind that F isn't differentiable at x = 1 or x = 2 (or at x = 0). Here's where you might start thinking about the Fund. Theorem of Calculus.

25. May 6, 2010

### mathgal

But doesn't the Fundamental Theorem of Calculus tell us that since our F is continuous everywhere, that it should be differentiable there? How do I show that x=1,2 are not differentiable? I know if the limit from the left and right are equal, it is continuous at that point.