1. ## monotone increasing

Let f : [a, b] -> R be a positive continuous function, i.e., f(x) >= 0 for all x in [a, b].

I want to show that the function F : [a, b] -> R defined as F(x) = integral of f with limits x and a, for x in [a, b] is monotone increasing (i.e., F(x) >= F(y) if x >= y).

any ideas? thankz

2. Originally Posted by dopi
Let f : [a, b] -> R be a positive continuous function, i.e., f(x) >= 0 for all x in [a, b].

I want to show that the function F : [a, b] -> R defined as F(x) = integral of f with limits x and a, for x in [a, b] is monotone increasing (i.e., F(x) >= F(y) if x >= y).

any ideas? thankz
Okay, you have the function
$F(x)=\int_x^a f(t) dt$
It exists since all continous functions are integrable.
Now, what ails me is that the upper limit is $a$, I think you wanted to write $c\in [a,b]$ thus hence the function,
$F(x)=\int_x^c f(t)dt=-\int_c^x f(t)dt$
The derivative of this function is (2nd fun. the. of cal.)
$F'(x)=-f(x)<0$
Thus the function is decreasing NOT increasing.

Thus, I think you wanted to write,
$F(x)=\int_c^x f(t)dt$
Then using the above concepts it is increasing on $[a,b]$

3. ## verifying

Originally Posted by ThePerfectHacker
Okay, you have the function
$F(x)=\int_x^a f(t) dt$
It exists since all continous functions are integrable.
Now, what ails me is that the upper limit is $a$, I think you wanted to write $c\in [a,b]$ thus hence the function,
$F(x)=\int_x^c f(t)dt=-\int_c^x f(t)dt$
The derivative of this function is (2nd fun. the. of cal.)
$F'(x)=-f(x)<0$
Thus the function is decreasing NOT increasing.

Thus, I think you wanted to write,
$F(x)=\int_c^x f(t)dt$
Then using the above concepts it is increasing on $[a,b]$
With this question that i posted i denoted F(x) = integral of f with upper limit x and lower limt alpha, but you have added (t) as in f(t), where as iv just wrote f, so i think it might be an riemann integra....does this affect the ansower of the question to what you have done?
thankz