Math
Estimation of Riemann integral over subinterval
Step-by-step mathematics solution: Estimation of Riemann integral over subinterval
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1. What the problem is asking (in plain language)
We are given a real‑valued function
[ f:[0,L]\longrightarrow \mathbb R , \qquad\text{which is Riemann‑integrable on }[0,L]. ]
Its total integral is
[ I=\int_{0}^{L}f(x)\,dx . ]
The average value of (f) on the whole interval is
[ \mu=\frac{I}{L}. ]
For a fixed length (r\;(0<r\le L)) we look at all sub‑intervals of that length,
([a,a+r]\subset[0,L]) (so (a\in[0,L-r])).
For each such sub‑interval we can compute the integral
[ \int_{a}^{a+r}f(x)\,dx . ]
The question is:
For which positive numbers (r) can we be sure (no matter which integrable (f) we start with) that there is at least one sub‑interval of length (r) whose integral equals (\displaystyle\frac{Ir}{L}= \mu r)?
In other words, for which (r) does the equality
[ \int_{a}^{a+r} f(x)\,dx = \frac{Ir}{L} \tag{1} ]
hold for some (a\in[0,L-r]) for every integrable (f)?
2. Proof – every (r\in(0,L]) works
2.1 A convenient auxiliary function
Define the cumulative integral (also called the primitive)
[ F(x)=\int_{0}^{x}f(t)\,dt ,\qquad x\in[0,L]. ]
Because (f) is Riemann‑integrable, (F) is continuous on ([0,L]).
Now introduce
[ \Phi(x)=F(x)-\mu x\;=\; \int_{0}^{x}f(t)\,dt-\frac{I}{L}x . \tag{2} ]
(\Phi) is continuous as a difference of continuous functions and satisfies
[ \Phi(0)=0,\qquad \Phi(L)=\int_{0}^{L}f(t)\,dt-\frac{I}{L}L=0 . \tag{3} ]
Thus (\Phi) starts and ends at the same value.
2.2 Sliding‑window function
For a fixed length (r\;(0<r\le L)) define
[ G_{r}(a)=\Phi(a+r)-\Phi(a),\qquad a\in[0,L-r]. \tag{4} ]
Using (2),
[ \begin{aligned} G_{r}(a) &= \Bigl(\int_{0}^{a+r}f-\mu (a+r)\Bigr)-\Bigl(\int_{0}^{a}f-\mu a\Bigr)\[2mm] &= \int_{a}^{a+r}f(t
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