# A problem involving Riemann Integrals

1. Oct 22, 2011

### fylth

I've been having some trouble with a maths problem and I hoped someone might be able to help.

We don't seem to have been taught most of what we need to do this, I understand Riemann integrals but what we've been taught and what they're asking for is just different.

I could do with a couple of hints or steps to get me in the right direction, it seems we've been taught different stuff to this, or stuff completely unrelated

Any help at all would be greatly appreciated

2. Oct 22, 2011

### Dickfore

The function:
$$\Psi(x, \epsilon) \equiv \left( \epsilon \, \vert f(x) \vert - \vert g(x) \vert \right)^2 \ge 0$$
is non-negative for all values of $x$ and $\epsilon$. But, if you expand it as a polynomial in $\epsilon$, you will get a quadratic function:
$$\psi(x, \epsilon) = \epsilon^2 \vert f(x) \vert^2 - 2 \epsilon \, \vert f(x) \vert \, \vert g(x) \vert + \vert g(x) \vert^2 \ge 0$$
But, for $\epsilon > 0$, we have the inequality
$$\frac{\epsilon}{2} \, \vert f(x) \vert^2 + \frac{1}{2 \epsilon} \, \vert g(x) \vert^2 \ge \vert f(x) g(x) \vert, \; \forall x \in \mathbb{R}$$
Taking $\epsilon \rightarrow 2\epsilon$ would lead you to an inequality that is very useful for you

3. Oct 22, 2011

### fylth

Thank you so much, that actually looks like maths I can understand, not the unrelated gibberish our lecturers seem to enjoy telling us

It's now almost 1 in the morning and I've been doing maths for the last 6 hours pretty much, I may be back tomorrow with a question about Holders Inequalities (same situation, my lecture notes dont match the questions in the slightest), but i'll have a better look when i've had some sleep

Thanks again!

4. Oct 23, 2011

### fylth

Hey, I have another problem using Riemann Integrals:

In the case of part (ii), we haven't actually been taught what the symbol of the integral with the line through means or used it in any way

Again, any help or just a nudge in the right direction would be much appreciated, this module has been great for giving us questions unrelated to what we've been taught

5. Oct 23, 2011

### Dickfore

Let's focus on the first problem. Did you solve it? If yes, post the solution here.

6. Oct 23, 2011

### fylth

Thats a good idea, here's what i've got (I'm no good with Latex so ill write it out longhand, say if anything is unclear)

Given
a
∫ (ε|f(x)| - |g(x)||2dx≥0 for all x, ε,
b

We can expand this to:
a
∫ε2|f(x)|2-ε|f(x)||g(x)|-ε|f(x)||g(x)|+|g(x)|2dx
b

=
a
∫ε2|f(x)|2-2ε|f(x)||g(x)|+|g(x)|2dx
b

So we know:
a
∫ε2|f(x)|2-2ε|f(x)||g(x)|+|g(x)|2dx ≥ 0
b

then
a
∫ε2|f(x)|2+|g(x)|2dx ≥
b

a
∫2ε|f(x)||g(x)|dx
b

Then dividing by 2ε:
a
∫(ε/2)|f(x)|2+(1/2ε)|g(x)|2dx
b

a
∫|f(x)||g(x)|dx
b

We now need to substitute ε by 2ε:
a
∫ε|f(x)|2+(1/4ε)|g(x)|2dx
b

=
a
∫ε|f(x)|2dx +
b

a
∫(1/4ε)|g(x)|2dx ≥
b

a
∫|f(x)||g(x)|dx
b

=a
ε∫|f(x)|2dx +
_b

_____a
(1/4ε)∫|g(x)|2dx ≥
_____b

a
∫|f(x)g(x)|dx
b

or
a
∫|f(x)g(x)|dx ≤
b

=a
ε∫|f(x)|2dx +
_b

_____a
(1/4ε)∫|g(x)|2dx
_____b

as required.

Sorry that's such a mess, must teach myself how to represent the symbols using latex :P

Last edited: Oct 23, 2011
7. Oct 23, 2011

### HallsofIvy

Staff Emeritus
That's because it is not a standard symbol. It is defined in this problem: it means "the integral of f from a to b divided by b- a" just as you are told.