True/false question on integration

  • #1
217
0
Let f(x) > 0 be a continuous function. Then F(x) = ∫f(t) dt ≥ 0 [0, x] for all values of x.

I would have chosen true, but the answer is false. If f(x) is always positive, then how can the integral ever be negative?

Thanks.
 

Answers and Replies

  • #2
13,049
6,931
is this answer in a book or the prof said so?

is the t going from 0 to x or x to 0? that could be the difference.
 
  • #3
217
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It is part of an old exam, the solutions say that the answer is false.
 
  • #4
13,049
6,931
then its probably due to whether its evaluated from 0..x or x..0
 
  • #5
217
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The bottom limit on the integral is 0 and the top limit is x.
 
  • #7
217
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We haven't discussed the differences between real and complex integrals, so it's real.
 
  • #8
13,049
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I can't think of any other trick so I think you're right and I'd go with that. You could ask a teacher about it.

For f(x)>0 then conceptually sum of the vertical strip areas from 0 to x would make the answer positive. The trick is looking for some way to have ( F(x) - F(0) ) < 0 but ( F(x) - F(0) ) is the area vertical strip areas from 0 to x.

So given that f(x) is real then the only way is for the integral to be negative is if its evaluated from x to 0, Since it wasn't explicitly shown at least in your example.
 
  • #9
217
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I think I may have gotten it. Is it because x can be any value, so that the integral can go from 0 to -3, for example, where we would receive a negative value?
 
  • #10
13,049
6,931
perhaps but wouldn't the closed interval be written as [x,0] to imply that x<0
 
  • #11
Bacle2
Science Advisor
1,089
10
If you're going in the standard direction [0,x] , then your F(x) is positive:

F(x):=∫x0 f(t)dt

So F(0)=0 ; F'(x)=f(x) , by FTCalculus, and f(x)>0 by assumption.

If x<0 , then F(x)<0 by same argument.
 

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