# Proof of the Mean Value Theorem for Integrals

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 $$(a,b)$$.