Let \(f\) be a continuous real-valued function on \([0,1]\). Prove that there exists at least one point \(\xi\in[0,1]\) such that \(\int_0^1 x^4 f(x)\,dx=\frac{1}{5}f(\xi)\).
Proof. Assume that \(f\) is a continuous real-valued function on \([0,1]\). Then, by the Intermediate Value Theorem we have that \(f\) attains its maximum and minimum on \([0,1]\). That is, for some \(a,b\in[0,1]\),
\[f(a)=\min\limits_{[0,1]}f(x) \qquad \text{and} \qquad f(b)=\max\limits_{[0,1]}f(x).\]
We now have \(f(a)\leq f(x)\leq f(b)\) for all \(x\in[0,1]\). This gives \[f(a)\int_0^1 x^4dx\leq \int_0^1 x^4f(x)dx\leq f(b)\int_0^1 x^4dx.\]
By the Fundamental Theorem of Calculus we know that
\[\int_0^1x^4dx=\frac{1}{5}.\]
Thus, it follows that
\[\frac{1}{5}f(a)\leq\int_0^1 x^4f(x)dx\leq \frac{1}{5}f(b)\] giving
\[f(a)\leq 5\int_0^1 x^4f(x)dx\leq f(b).\]
By the Intermediate Value Theorem, there exists \(\xi\in[0,1]\) such that
\[f(\xi)=5\int_0^1 x^4f(x)dx.\]
Therefore, we have that there exists \(\xi\in[0,1]\) such that \(\int_0^1 x^4 f(x)dx=\frac{1}{5}f(\xi)\). ◻
Solution
1. Let \(m = \min_{[0,1]} f\) and \(M = \max_{[0,1]} f\).
\(f\) is continuous on the compact set \([0,1]\), so it attains its extrema.
2. \(m/5 \le \int_0^1 x^4 f(x)\,dx \le M/5\).
\(m \le f(x) \le M\) for all \(x\), and \(x^4 \ge 0\), so \(m x^4 \le x^4 f(x) \le M x^4\); integrating and using \(\int_0^1 x^4\,dx = 1/5\) gives the claim.
3. \(5\int_0^1 x^4 f \in [m, M] = f([0,1])\).
by 2, \(5\int_0^1 x^4 f \in [m, M]\).
4. There is \(\xi \in [0,1]\) with \(\int_0^1 x^4 f(x)\,dx = \frac15 f(\xi)\).
by 3, \(5\int_0^1 x^4f = f(\xi)\) for some \(\xi \in [0,1]\).
5. Q.E.D.