# Thread: integrals of nonpositive functions

1. ## integrals of nonpositive functions

show that if "f" is integrable then

f(x) < or equal to 0 on [a,b]

is greater than or = to zero

2. Originally Posted by 10tomlinsonb
show that if "f" is integrable then

f(x) < or equal to 0 on [a,b]

is greater than or = to zero
Well that is not true.
It is true that $\int_a^b {f(x)dx} \leqslant 0$ .

3. sorry that is what i meant to write. can you show me why that is true?

4. Originally Posted by 10tomlinsonb
sorry that is what i meant to write. can you show me why that is true?
That depends upon who much you know.
Is this a problem in Riemann integrals?

5. it is a problem in definite integrals

6. Originally Posted by 10tomlinsonb
it is a problem in definite integrals
I am not sure what that implies.

In most basic calculus textbooks there is a theorem that states:
If $\left( {\forall x \in \left[ {a,b} \right]} \right)\left[ {f(x) \leqslant g(x)} \right]$ then it is true that $\int_a^b {f(x)dx} \leqslant \int_a^b {g(x)dx}$.
if your text has that theorem then let $g(x)=0$ and use the theorem.
Otherwise, you need to use upper sums.

7. can you show me how you would prove this using upper sums

8. please help me prove this because i am so confused and i need it for a homework problem

9. The reason Plato asked if you knew about Riemann sums is that how you prove such a basic property of integrals depends upon how you define integrals in the first place! And "Riemann sums" is a common method of defining integrals. You mention of "upper sums" indicates that's what you have learned, although "upper sums" is only a very small part of that.

Given that f(x) is integrable on [a, b], then for any partition of [a, b] (dividing it into smaller subintervals), choosing any $x_i$ in the $i^{th}$ interval, with width $\Delta x_i$, the Riemann sum is $\sum f(x_i)\Delta x_i$ and, as we take "finer and finer" partitions, meaning that the largest $\Delta x_i$ goes to 0 as we take more and more intervals in the partition, and that sum goes to some fixed value, I, whic is the integral, $\int_a^b f(x)dx$.

Now, if $f(x)\le 0$ for all x in [a, b], in particular, $f(x_i)\le 0$ for every partition and for every $x_i$ in a partition. That, in turn, means that $f(x_i)\Delta x_i\le 0$ for every partition and every i in a partition, and so $\sum f(x_i)\Delta x_i\le 0$ for every partition. The integral, the limit of negative numbers can't be positive: $\int_a^b f(x)dx\le 0$.

10. my teacher also told that

(b-a) is max of f(x) therefore,

≤ (b-a) ≤ 0

can you expain why this is

11. Originally Posted by 10tomlinsonb
sorry that is what i meant to write. can you show me why that is true?
A single counter example will show this: let b>a, and f(x)=-1 and you can now show that the integral is negative.

CB