My Single Variable Calc Textbook asked me to prove the Mean Value Theorem for Integrals by applying the Mean Value Theorem for Derivatives to the function $F(x)=\displaystyle\int_a^xf(t)dt$. I'm pretty sure that my proof is correct, but a correct proof is not listed in the textbook. Also, it would be helpful to critique my proof's format because I've been trying to format my proofs more professionally.

$$\text{Theorem: If $f$ is continuous on [a,b], then there exists a number $c$ in [a,b] such that}$$ $$f(c)(b-a)=\int_a^b f(t)dt$$ $$\text{Proof:}$$ $$F(x)=\int_a^xf(t)dt$$ $$\text{By the Fundamental Theorem of Calculus, we have}$$ $$F'(x)=f(x)$$ $$\text{By the Mean Value Theorem for Derivatives}$$ $$F'(c)=\frac{F(b)-F(a)}{b-a}$$ $$f(c)=\frac{F(b)-F(a)}{b-a}$$ $$f(c)(b-a)=F(b)-F(a)$$ $$\text{By the Fundamental Theorem of Calculus}$$ $$f(c)(b-a)=\int_a^bf(t)dt$$


The proof is correct, but can be made shorter and more accurate (there should be “there exists $c$” somewhere).

Consider $F(x)=\int_a^x f(t)\,dt$; then, by the fundamental theorem of calculus, $F'(x)=f(x)$, for every $x\in[a,b]$; moreover, $F(b)-F(a)=\int_a^b f(t)\,dt$.

Since $F$ is continuous over $[a,b]$ and differentiable over $(a,b)$, the mean value theorem applies and there exists $c\in(a,b)$ such that $$ \frac{F(b)-F(a)}{b-a}=F'(c) $$ that is, $$ \int_a^b f(t)\,dt=(b-a)f(c) $$

By the way, the proof can be given without mentioning the mean value theorem. Since $f$ is continuous over the interval $[a,b]$, it has a maximum value $M$ and a minimum value $m$. Then, by definition of integral and from $m\le f(t)\le M$, we have $$ m(b-a)\le\int_a^b f(t)\,dt\le M(b-a) $$ By the intermediate value theorem, there exists $c\in[a,b]$ such that $$ f(c)=\frac{1}{b-a}\int_a^b f(t)\,dt $$ This is somewhat less precise than the other version, because we cannot state, without further work, that $c$ can be chosen in $(a,b)$.


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