Comparison of Riemann integral to accumulation function

[TopCat]

Let f:[0,1]→ℝ be an increasing function. Show that for all x in (0,1],

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


So by working backwards I got to trying to show that (1-x)\int_{0}^{1}f (t) \,dt \le \int_{x}^{1}f (t) \,dt. 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.

 

[lanedance]

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
(1-x)c&lt;\int_x^1 f(t) dt
\int_0^x f(t) dt &lt; xc

the second gives
\frac{1}{x}\int_0^x f(t) dt &lt; c

 

[lanedance]

also it may help to think of it like this

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

\int_{0}^{1}f (t) \,dt 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

 

[TopCat]

Thanks for the intuition. Now I was able to easily get the inequality

\frac{1}{x}\int^{x}_{0}f\,dt \le \frac{1}{1-x}\int^{1}_{x}f\,dt

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