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
$$(1-x)c<\int_x^1 f(t) dt$$
$$\int_0^x f(t) dt < xc$$

the second gives
$$\frac{1}{x}\int_0^x f(t) dt < c$$

 

[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. :(