Question

Can anyone help me think about a way to prove or disprove this? If f(x) >=0 and b >=0, then b*(integral from b to infinity of f(x)dx) <= (integral from b to infinity of x*f(x)dx).

Answer 1

how about f(x)= e^-x
b=1?

Answer 2

nope that does not work

Answer 3

seems rather unlikely to me

Answer 4

try \(f(x)=\frac{1}{x^2}\)

Answer 5

\[ \large
b\int_b^\infty f(x)dx \leq \int_b^\infty xf(x)dx
\]This?

Answer 6

ya thats right wio

Answer 7

then
\[\int_b^\infty\frac{dx}{x^2}\] is finite, whereas
\[\int_b^{\infty}\frac{dx}{x}\] is not

Answer 8

oh i forgot the \(b\) out front
no matter, take \(b=1\)

Answer 9

right i saw that and was not sure if that would count as an answer. I then tried f(x) =
1/x^3

Answer 10

all you need is one example to disprove, right?

Answer 11

ok so because the first one is finite while the other one is infinite, that disproves the inequality?

Answer 12

Since \(b\) is positive: \[
\large
\int_b^\infty f(x)dx \leq \int_b^\infty \frac{x}{b}f(x)dx
\]

Answer 13

Maybe then try integration by parts to see if you can get something interesting.

Answer 14

yes one is finite and the other is infinite
that certainly disproves the inequality

Answer 15

alright cool! Thank you i really appreciate it

Answer 16

@satellite73 Really? So \(6 < \infty\) for example, is false... can be used as a counter example?

Answer 17

actually it is a rather silly inequality, if it is what i read it to be
if you take \(b=1\) then why on earth would
\[\int_1^{\infty}f(x)dx\] be less than \[\int_1^{\infty}xf(x)dx\]?
try with \(f(x)=\frac{1}{x^3}\) or anything of that type

Answer 18

oh damn hold the phone i am way way wrong!!!

Answer 19

ignore everything i wrote
everything

Answer 20

To me it looked like is was pretty right on.

Answer 21

no i had the whole thing backwards. now i have to think
maybe it is true

Answer 22

after all, as \(x\to \infty\) at some point \(x>b\)

Answer 23

Suppose \(c = x-b\) Then we have \(x = b+c\).

Answer 24

\[
\large
b\int_b^\infty f(x)dx \leq \int_b^\infty (b+c)f(x)dx
\]
This gets us:
\[
\large
b\int_b^\infty f(x)dx \leq b\int_b^\infty f(x)dx +c\int_b^\infty f(x)dx
\]Then:
\[\large
0 \leq c\int_b^\infty f(x)dx
\]

Answer 25

oops, let me redo that

Answer 26

\[
\large
b\int_b^\infty f(x)dx \leq b\int_b^\infty f(x)dx +\int_b^\infty (x-b) f(x)dx
\]

Answer 27

Since we start integrating at \(b\), it's clear that \(x-b\) is positive and \(f(x)\) is positive.

Answer 28

So we could say \[ \large
0 \leq \int_b^\infty (x-b)f(x)dx
\]

Answer 29

why can we say that x-b is positive
?

Answer 30

perhaps I should be saying 'non negative' rather than positive.

Answer 31

Remember that an integral is a infinite sum. The sum is starting with \(x=b\) and ending with \(x=\infty\).

Answer 32

ahhhh right ok

Answer 33

You'll probably want to convert the integral to summation notation to show this.

Answer 34

\[ \large
\int_b^\infty (x-b)f(x)dx = \lim_{n \to \infty} \sum_i^n (x_i^*-b)f(x_i^*)\Delta x
\]

Answer 35

\[
b \leq x_i^* < \infty
\]

Answer 36

\[
0 \leq x_i^* -b < \infty
\]

Answer 37

ok i think the summation notation is making it a little harder for me to understand

Answer 38

i understand the whole integral part from above. is that good to show that it is true?

Answer 39

it seems like it should be but im kind of new to proofs

Answer 40

Ummm, I think it would be a bit iffy, but I think it should be okay.

Answer 41

It's a matter of rigor. Perhaps I'm being too rigorous.

Answer 42

alright i know what you mean

Answer 43

its starting to make sense

Answer 44

Basically I'm showing that you're adding a bunch of non-negative numbers, resulting in a non-negative number.

Answer 45

Basically this proof will be doing everything we did here backwards.

Answer 46

and because they can not be negative then if you multiply by x there is no way it will be less then if you dont multiply by x right?

Answer 47

You want to start by asserting this: \[
\large
0 \leq \int_b^\infty (x-b)f(x)dx
\]THEN you add \[
\large
b\int_b^\infty f(x)dx
\] To both sides.

Answer 48

So that first assertion is key to proving it.