Comparison of Riemann integral to accumulation function

  • Thread starter TopCat
  • Start date
  • #1
58
0
Let f:[0,1]→ℝ be an increasing function. Show that for all x in (0,1],

[tex]\frac{1}{x}\int_{0}^{x}f (t) \,dt \le \int_{0}^{1}f (t) \,dt [/tex]


So by working backwards I got to trying to show that [itex](1-x)\int_{0}^{1}f (t) \,dt \le \int_{x}^{1}f (t) \,dt [/itex]. While I know both sides are equal at x=1, the LHS is decreasing, and the RHS is continuous, I haven't been able to find a reason to say the inequality is true (probably nothing there).

Should I pass to considering upper and lower sums, or is there some fancy way to use integration by parts here? I haven't come across anything compelling in all my random scratch work.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
how about something like this...

f is increasing so
f(a)<f(b) for all a<b (strictly increasing?)

now say f(x) = c, can you show
[tex](1-x)c<\int_x^1 f(t) dt [/tex]
[tex]\int_0^x f(t) dt < xc [/tex]

the second gives
[tex]\frac{1}{x}\int_0^x f(t) dt < c [/tex]
 
Last edited:
  • #3
lanedance
Homework Helper
3,304
2
also it may help to think of it like this

[itex] \frac{1}{x}\int_{0}^{x}f (t) \,dt [/itex] is the average value of the function over the interval [0,x]

[itex]\int_{0}^{1}f (t) \,dt [/itex] is the average value of the function over the interval [0,1]

As the function is increasing, so will its average increase over a larger interval
 
  • #4
58
0
Thanks for the intuition. Now I was able to easily get the inequality

[tex]\frac{1}{x}\int^{x}_{0}f\,dt \le \frac{1}{1-x}\int^{1}_{x}f\,dt[/tex]

but I'm not sure how that can show the LHS ≤ [itex]\int^{1}_{0}f\,dt[/itex]. I'm sure I'm being blind and missing something super obvious. :(
 

Related Threads on Comparison of Riemann integral to accumulation function

Replies
1
Views
2K
Replies
11
Views
2K
Replies
1
Views
2K
Replies
0
Views
903
Replies
4
Views
2K
Replies
12
Views
9K
Replies
1
Views
477
Replies
6
Views
1K
Replies
10
Views
8K
Replies
22
Views
3K
Top