# Integral Proof

• May 14th 2010, 02:15 AM
olski1
Integral Proof
I have been asked to justify if the following statement is true or false.

'If the integral of f(x) >= 0 on [a,b], then f(x) is >= 0 on [a,b].

Would (8-x^2) over the domain [-4,4] be a counter example to this? thus justifying that the statement is false.

I beleive so as f(x) is <= 0 for some values but the area over that domain is positive. Can anyone tell me if iam on the right path?
• May 14th 2010, 02:25 AM
Prove It
Quote:

Originally Posted by olski1
I have been asked to justify if the following statement is true or false.

'If the integral of f(x) >= 0 on [a,b], then f(x) is >= 0 on [a,b].

Would (8-x^2) over the domain [-4,4] be a counter example to this? thus justifying that the statement is false.

I beleive so as f(x) is <= 0 for some values but the area over that domain is positive. Can anyone tell me if iam on the right path?

$\displaystyle 8 - x^2$ is NOT $\displaystyle \geq 0$ for the entire domain $\displaystyle [-4, 4]$.
• May 14th 2010, 02:45 AM
olski1
So doesnt that proove that the statement is false?
• May 14th 2010, 02:55 AM
Prove It
Quote:

Originally Posted by olski1
So doesnt that proove that the statement is false?

Definitely not.

I think it should be obvious that if a function is nonnegative for an entire region, then the area between the function and the $\displaystyle x$ axis can never fall below the $\displaystyle x$ axis.
• May 14th 2010, 03:29 AM
Defunkt
Quote:

Originally Posted by olski1
I have been asked to justify if the following statement is true or false.

'If the integral of f(x) >= 0 on [a,b], then f(x) is >= 0 on [a,b].

Would (8-x^2) over the domain [-4,4] be a counter example to this? thus justifying that the statement is false.

I beleive so as f(x) is <= 0 for some values but the area over that domain is positive. Can anyone tell me if iam on the right path?

Your counter example is correct, since $\displaystyle \int_{-4}^4 8-x^2 ~ dx = 8\int_{-4}^4 dx ~ - \int_{-4}^4 x^2 ~ dx = 8\cdot 8 - (\frac{(4)^3}{3} - \frac{(-4)^3}{3}) = 64 - \frac{2}{3}64 = \frac{64}{3} > 0$

however $\displaystyle f(3) = -1 < 0$

Quote:

Definitely not.

I think it should be obvious that if a function is nonnegative for an entire region, then the area between the function and the axis can never fall below the axis.
I think you were thinking of the other direction of the question.