Prove Two Derivatives are Equal at a Point

Question

Suppose $f$ and $g$ are continuous on $[a,b]$ and $\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx$. Prove that there is $c\in [a,b]$ such that $f(c)=g(c)$.

So, I've had a tough time with this problem. I've looked at trying to prove by contradiction, or by contrapositive, and even directly. Graphically, we understand this can be true in two cases, either $f(x)=g(x)$, or on some fraction of domain, say, $[a,c]$ that $f(x)\geq g(x)$ or $g(x)\geq f(x)$, then on $[c,b]$ either $f(x)\geq g(x)$ or $g(x)\geq f(x)$.

Dividing this $[a,b]$ into a partition basically, can ensure that these function lines intersect at some point $(c,f(c))$ and that their corresponding areas for one function above another on each side of $c$ are equal. I don't know where I am to go from here. I don't think there's any useful theorems to use. I kind of see how it might be a Rolle's Theorem, IVT, or MVT, but I'm not entirely sure.

Answer 1

Note that $h(x) = \int_a^x f(t) - g(t) \,\mathrm{d}t$ is zero when $x = a$ and when $x = b$. By the mean value theorem, there is a $c \in (a,b)$ where $h'(c) = 0$. By the Fundamental Theorem of Calculus (part 2), $h'(x) = f(x) - g(x)$, so $0 = h'(c) = f(c) - g(c)$.

Answer 2

Integral Mean Value Theorem applied to \begin{align*} (f-g)(c)=\dfrac{1}{b-a}\int_{a}^{b}(f-g)(x)dx=0. \end{align*}

Answer 3

If $f(c) \ne g(c)$ for all $c$ then we must have $f(c) > g(c)$ for all $c$ (or the other way around).

Then we must have $\int f > \int g$.

Answer 4

Suppose one of the antiderivatives of f(x) is F(x), and one of the antiderivatives of g(x) is G(x). Then F(b)-F(a)=G(b)-G(a) with F and G continuous on [a,b] and differentiable on (a,b). By Cauchy's mean value theorem, $$1=\frac{F(b)-F(a)}{G(b)-G(a)}=\frac{F'(\xi)}{G'(\xi)}=\frac{f(\xi)}{g(\xi)}$$ for some $\xi\in(a,b)$. In this case, we have $f(\xi)=g(\xi)$, hence proven.