- #1

IntegrateMe

- 217

- 1

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.

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- Thread starter IntegrateMe
- Start date

- #1

IntegrateMe

- 217

- 1

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

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

- #3

IntegrateMe

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

- #4

jedishrfu

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

- #5

IntegrateMe

- 217

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

- #6

jedishrfu

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okay so is f(x) real or complex?

- #7

IntegrateMe

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

- #8

jedishrfu

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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

IntegrateMe

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- #10

jedishrfu

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

- #11

Bacle2

Science Advisor

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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.

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