Mean Value Theorem of Integration
Continuous functions attain their mean value.
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Study Mean Value Theorem of Integration
Now we present the mean value theorem of integration and its manifold consequences.
Theorem 1 (Mean Value Theorem of Integration). Let \(f,g:[a,b]\to\mathbb{R}\) be continuous and let \(g(x)\geq0\) for all \(x\in[a,b]\) (i.e., \(g\geq0\)). Then there exists some \(\hat{x}\in[a,b]\) such that \[\int_a^bf(x)g(x)\, dx=f(\hat{x})\cdot \int_a^bg(x)\, dx.\]
Proof: Let \[m=\min\{f(x)\::\:x\in[a,b]\},\qquad M=\max\{f(x)\::\:x\in[a,b]\}.\] Since \(g\geq0\), for all \(x\in[a,b]\) holds \(mg(x)\leq f(x)g(x)\leq Mg(x)\) and, by the monotonicity of the integral holds \[m\int_a^bg(x)\, dx=\int_a^bmg(x)\, dx\leq \int_a^bf(x)g(x)\, dx\leq \int_a^bMg(x)\, dx\leq M\int_a^bg(x)\, dx.\] Then there exists some \(m\leq\mu\leq M\) such that \[\mu\int_a^bg(x)\, dx=\int_a^bf(x)g(x)\, dx\] Since \[\min\{f(x)\::\:x\in[a,b]\}\leq\mu\leq\max\{f(x)\::\:x\in[a,b]\},\] the intermediate value theorem implies that there exists some \(\hat{x}\in[a,b]\) with \(\mu=f(\hat{x})\) and thus \[f(\hat{x})\int_a^bg(x)\, dx=\int_a^bf(x)g(x)\, dx.\] \(\Box\)
Let \(f:[a,b]\to\mathbb{R}\) be continuous. Then there exists some \(\hat{x}\in[a,b]\) such that \[\int_a^bf(x)\, dx=f(\hat{x})\cdot (b-a).\]
Proof: Apply the mean value theorem to \(g=1\). Observing that \[\int_a^bg(x)\, dx=\int_a^b1\, dx= b-a,\] the result then follows immediately. \(\Box\)
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