Question

Help with proof

Answer 1

@zepdrix

Answer 2

Is our goal to use Reimans sum at all? I know this is essentially part of the fundamental theorem of calculus

Answer 3

if you can't use the obvious, then i guess riemann sum is the way to go

Answer 4

@satellite73 So would we start by stating the sum?

Answer 5

start with dividing the interval into 1/n widths called Δx and then state that f(xi) Δx < g(xi) Δx

Answer 6

the the sum on the left is less then the sum on the right.

Answer 7

that's just the outline though.

Answer 8

Hi! not to interup you getting help, but im really in the mood to fan and give out medals! Does anyone want to help me out with this problem?

Answer 9

@Loser66 can you talk me through this proof?

Answer 10

oh, which one should I help? @redheadangel or @DarkBlueChocobo ?

Answer 11

:p *raises hand*

Answer 12

hahaha... .but your problem is not easy. I will try. Give me few minuses

Answer 13

Thank you so much.

Answer 14

oh, I don't know why I see it too easy. hahaha... I might be wrong but it is so simple to me
\(g(x) \geq f(x)\\g(x) -f(x) \geq 0\)

Take integral both sides from a to b, we have

\[\int_a^b (g(x) -f(x) )dx =\int_a^b g(x) dx -\int_a^b f(x) dx \geq 0\]

That shows
\[\int_a^b g(x) dx \geq \int_a^b f(x) dx\]

Answer 15

Can you talk me through how you viewed this?

Answer 16

I look at it and I can solve it by plugging in values that demonstrate its true, but I do not know how to do it generally.

Answer 17

It is trivial, right?