A problem involving Riemann Integrals

In summary: Let's focus on the first problem. Did you solve it? If yes, post the solution here.Sorry, I don't understand how to post a solution.
  • #1
fylth
4
0
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.

2iuazgx.png


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
 
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  • #2
The function:
[tex]
\Psi(x, \epsilon) \equiv \left( \epsilon \, \vert f(x) \vert - \vert g(x) \vert \right)^2 \ge 0
[/tex]
is non-negative for all values of [itex]x[/itex] and [itex]\epsilon[/itex]. But, if you expand it as a polynomial in [itex]\epsilon[/itex], you will get a quadratic function:
[tex]
\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
[/tex]
But, for [itex]\epsilon > 0[/itex], we have the inequality
[tex]
\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}
[/tex]
Taking [itex]\epsilon \rightarrow 2\epsilon[/itex] would lead you to an inequality that is very useful for you
 
  • #3
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 don't match the questions in the slightest), but i'll have a better look when I've had some sleep

Thanks again!
 
  • #4
Hey, I have another problem using Riemann Integrals:

2k1sti.png


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
Let's focus on the first problem. Did you solve it? If yes, post the solution here.
 
  • #6
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:
  • #7
fylth said:
Hey, I have another problem using Riemann Integrals:

2k1sti.png


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
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.

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
 

What is a Riemann Integral?

A Riemann Integral is a mathematical concept that is used to find the area under a curve. It is named after the mathematician Bernhard Riemann and is an important tool in calculus.

How do you solve a problem involving Riemann Integrals?

To solve a problem involving Riemann Integrals, you must first determine the limits of integration and then break the area under the curve into smaller, manageable pieces. You then use a Riemann sum to approximate the area and take the limit as the number of pieces approaches infinity to find the exact value of the integral.

What is the difference between a definite and indefinite Riemann Integral?

A definite Riemann Integral has specific limits of integration and gives a numerical value as the result. An indefinite Riemann Integral does not have specific limits of integration and gives a function as the result, which can then be evaluated at specific points.

What is the importance of Riemann Integrals in mathematics?

Riemann Integrals are important in mathematics because they allow us to find the area under a curve, which is crucial in many real-world applications. They also help us understand the behavior of functions and are fundamental in the development of calculus.

Are there any limitations to using Riemann Integrals?

Yes, there are some limitations to using Riemann Integrals. They can only be used for functions that are continuous on a closed interval. They also may not work for some non-standard functions, such as those that are discontinuous or non-differentiable.

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