True or False: With Integrals

  • #1
RJLiberator
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True or False? Let a and b be real numbers, with a < b, and f a continuous function on the interval [a, b].

a) If a=b then [itex]\int^{b}_{a} f(x)dx = 0[/itex]

My answer: This is TRUE, because while this integral would have a height, it would NOT have a width and area being l*w will result in 0.

b) If [itex]a \neq b[/itex], then [itex]\int^{b}_{a}f(x)dx \neq 0[/itex]

My answer: This is FALSE, because there will exist a height and a width, however, half of the area can be negative area and half of the area can be positive area canceling each other out and creating 0 area

c) If [itex]a \neq b[/itex], then [itex]\int^{b}_{a}f(x)dx = 0, then f(x) = 0 for all x \in [a,b] [/itex]

My Answer: This is FALSE, however, I am having trouble finding an example.

D) If [itex]a \neq b[/itex], then [itex]\int^{b}_{a}|f(x)|dx = 0, then f(x) = 0 for all x \in [a,b] [/itex]

My Answer: EDIT: This is True.

Correct answers?

Thanks for checking in with me and guiding me.
 
Last edited:

Answers and Replies

  • #2
Seems correct to me. Why did you repeat two times the c question? As an example you could for istance think to a periodic function on a symmetric interval. Can you figure it out?
 
  • #3
RJLiberator
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Hi val, thanks for the help.

The difference between c and D is the absolute value of f(x).

Periodic function on a symmetric interval? Would sin(x) be one?
 
  • #4
Yes sin(x) integrated between -pi and pi is a perfect example for c. I didn't notice the absolute value, then are you sure about d?
 
  • #5
RJLiberator
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OH. Wow. I see it now more clearly.

The absolute value negates the situation. It turns the negative area into a positive area causing the statement to be true.

Thanks 0_0.
 
  • #6
RJLiberator
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Does anyone have any tips on proving these statements that I claim as true? More mathematically rigorous proofs.
 
  • #7
RJLiberator
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Okay, I am having a little problem with C and D.

If I use sin(x) and integrate it between -pi and pi, then I receive an area of 0. But what does it mean for f(x) = 0 for all x as a set of [a,b] ? Do I merely plug in a number from the set into f(x). Example: sin(-pi/2) = -1 to disprove this statement?

If so, how come in D it wouldn't work the same way?
 
  • #8
BiGyElLoWhAt
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Well let me throw something out for a just because it seems like you have a relatively descent idea of how to go about this. I hope this will give you an idea as to how to state d.

Let ##F(x) = \int f(x)dx##
then ##\int_{a}^b f(x)dx = F(b) - F(a)##

What happens if a = b?

Now for d. I first want to clarify your wording:
D) If [itex]a \neq b[/itex], then [itex]\int^{b}_{a}|f(x)|dx = 0 [/itex], then [itex]f(x) = 0 [/itex] for all [itex]x \in [a,b] [/itex]

Did you intend the bolded 'then' to be 'and'?

If so, state what you do know about ##|f(x)|## i.e. what do the | | symbols do to the possible outputs of f(x)?
 
  • #9
RJLiberator
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Hi Bigyellowhat,

Yes, I did intend for the bolded to be 'and.' My apologies.

The possible outputs of f(x) must be positive in this scenario.
So this would mean that the statement cannot exist? It must be false.
The only value that makes f(x)=0 in D is a=b and a cannot equal b.
 
  • #10
RJLiberator
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D is rather tricky. I understand it now, but if anyone has concrete examples, that would be great.

Essentially, I have to take the value of the integral of f(x)dx as equal to 0, even tho a cannot equal b.

Thus, the limited subset we are offered must always equal 0.

Interesting.
 
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  • #11
PeroK
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D is rather tricky. I understand it now, but if anyone has concrete examples, that would be great.

Essentially, I have to take the value of the integral of f(x)dx as equal to 0, even tho a cannot equal b.

Thus, the limited subset we are offered must always equal 0.

Interesting.

Note that for D to be true, f must be continuous. First, therefore, you might like to try to find a non-continuous f for which the integral is 0, but f is not identically 0.

To prove it for continuous f, you probably need to know the definition of continuity.

Note: it might simplify things a little just to assume that f(x) ≥ 0 and drop the modulus.
 

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