# integrals of nonpositive functions

• March 7th 2010, 12:16 PM
10tomlinsonb
integrals of nonpositive functions
show that if "f" is integrable then

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

http://upload.wikimedia.org/math/1/8...fb77870925.png is greater than or = to zero
• March 7th 2010, 12:29 PM
Plato
Quote:

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

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

http://upload.wikimedia.org/math/1/8...fb77870925.png is greater than or = to zero

Well that is not true.
It is true that $\int_a^b {f(x)dx} \leqslant 0$ .
• March 7th 2010, 12:36 PM
10tomlinsonb
sorry that is what i meant to write. can you show me why that is true?
• March 7th 2010, 12:44 PM
Plato
Quote:

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?
• March 7th 2010, 01:03 PM
10tomlinsonb
it is a problem in definite integrals
• March 7th 2010, 01:16 PM
Plato
Quote:

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.
• March 7th 2010, 01:29 PM
10tomlinsonb
can you show me how you would prove this using upper sums
• March 7th 2010, 02:08 PM
10tomlinsonb
please help me prove this because i am so confused and i need it for a homework problem
• March 7th 2010, 02:30 PM
HallsofIvy
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: $\int_a^b f(x)dx\le 0$.
• March 7th 2010, 02:57 PM
10tomlinsonb
my teacher also told that

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

http://www.mathhelpforum.com/math-he...8b16c3ed-1.gif ≤ (b-a) ≤ 0

can you expain why this is
• March 8th 2010, 04:41 AM
CaptainBlack
Quote:

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