# True/false question on integration

1. Feb 6, 2012

### IntegrateMe

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.

2. Feb 6, 2012

### Staff: Mentor

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. Feb 6, 2012

### IntegrateMe

It is part of an old exam, the solutions say that the answer is false.

4. Feb 6, 2012

### Staff: Mentor

then its probably due to whether its evaluated from 0..x or x..0

5. Feb 6, 2012

### IntegrateMe

The bottom limit on the integral is 0 and the top limit is x.

6. Feb 6, 2012

### Staff: Mentor

okay so is f(x) real or complex?

7. Feb 6, 2012

### IntegrateMe

We haven't discussed the differences between real and complex integrals, so it's real.

8. Feb 6, 2012

### Staff: Mentor

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. Feb 6, 2012

### IntegrateMe

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. Feb 6, 2012

### Staff: Mentor

perhaps but wouldn't the closed interval be written as [x,0] to imply that x<0

11. Feb 6, 2012

### Bacle2

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.