- #1

- 217

- 0

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter IntegrateMe
- Start date

- #1

- 217

- 0

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

jedishrfu

Mentor

- 13,049

- 6,931

is the t going from 0 to x or x to 0? that could be the difference.

- #3

- 217

- 0

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

- #4

jedishrfu

Mentor

- 13,049

- 6,931

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

- #5

- 217

- 0

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

- #6

jedishrfu

Mentor

- 13,049

- 6,931

okay so is f(x) real or complex?

- #7

- 217

- 0

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

- #8

jedishrfu

Mentor

- 13,049

- 6,931

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

- 0

- #10

jedishrfu

Mentor

- 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

F(x):=∫

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.

Share: