# True/false question on integration

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.

jedishrfu
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.

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

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

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

jedishrfu
Mentor
okay so is f(x) real or complex?

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

jedishrfu
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.

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?

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

Bacle2