I figure now to just subtract the left side from the right, setting the equation to f(x).
Then, take the first derivative of it, which is g(x) [2\int0^x g(t) dt - (g(x)2)]
To prove f' increases, use the fact that g'\geq 0, demonstrating g(x) increases.
So, it can be said that...
I found this problem elsewhere, so I was wondering how to prove a specific part of it to be positive, because my teacher wanted a better explanation of it.
http://www.mathlinks.ro/viewtopic.php?t=325915
My questions were posted in the forum there as well.
Also, I realize that I misread...
Oops. I misplaced my 2. Thanks.
So, I was thinking about the application of the MVT...
Any ideas, hints if I am in the right direction, because I did a similar problem that made me work in the opposite direction of this one, using MVT.
It read: let a,b be differentiable on \Re and suppose...
I am getting confused a bit.
Let's see, the derivative of the integral of g(x) dx is just g(x), right?
So, then, with the square on it is it, g2(x)g'(x)
?
Office_Shredder...
So, if I use a simple function of just x, then the derivative of it is...
3x^2 \leq (1)^2
which is not true, if my values for x \geq 0 .
Is that what you mean (did I do that right)? Is that the contradiction you were referring to, since that is true when x
\leq 0
?
Thank you Office_Shredder and Mark44.
Office_Shredder has it written correctly, the part about the parentheses is correct, but it is g(x) dx, not with t's. Though that shouldn't make a difference, right??
Mark44, thank you as well, your Latex writing of the problem is correct, although it...
Homework Statement
I apologize for not knowing how to use Latex, so I will type the problem as it is read...
Prove that for all x greater than or equal to 0, we have
the integral from 0 to x for [g(x)]^3 dx which is less than or equal to
(the integral from 0 to x for g(x) dx)^2...