Spivak, Ch. 13 "Integrals", Problem 29: Proof that $f$ integrable on $[a,b]$ $\implies \exists x, x \in [a,b] \land \int_a^x f = \int_x^b f$. 


*Suppose that $f$ is integrable on $[a,b]$. Prove that there is a number $x$ in $[a,b]$ such that $\int_a^x f=\int_x^b f$. Show by example that it is not always possible to choose $x$ to be in $(a,b)$.


Here is my solution
Suppose $f$ is integrable on $[a,b]$.
Let $F(x)=\int_a^x f(t)dt$.
Then (by a Theorem in the main text of Chapter 13 of Spivak's Calculus), $F$ is continuous on $[a,b]$.
Now, $F(a)=0$ and $F(b)=\int_a^b f(t)dt$.
Let's use a proof by cases argument.
Case 1: $\int_a^b f(t)dt \neq 0$
Then, by the Intermediate Value Theorem, there is some $x \in (a,b)$ such that $F(x)=\frac{\int_a^b f(t)dt}{2}$.
But this just means that $\int_a^x f(t)dt = \frac{\int_a^b f(t)dt}{2}$, and therefore $\int_a^x f(t)dt = \int_x^b f(t)dt$.
Case 2: $\int_a^b f(t)dt = 0$
I would like to confirm: we cannot apply the Intermediate Value Theorem here, correct?
If we choose $x=a$ then we have
$$\int_a^x f = \int_a^a f = 0 = \int_x^b = \int_a^b$$
We could also choose $x=b$ and we would have
$$\int_a^x = \int_a^b = 0 = \int_x^b=\int_b^b$$
Either choice gives us the desired result.
In both possible cases considered in the proof by cases, we reached the desired result.
Therefore, by proof by cases, we can assert
$$f \text{ integrable on } [a,b]\implies  \exists x, x \in [a,b] \land \left ( \int_a^x f = \int_x^b f \right )$$
$\blacksquare$
As for the second part of the question, all we need to do is find a situation that falls into case 2 in the proof above. For example a function such as
$$f(x)=c-\frac{2c}{b-a}(x-a)$$

My main question is: the proof of case 2 above didn't actually prove that there isn't an $x \in (a,b)$ such that $\int_a^x f = \int_x^b f$, correct? It just proved that $a$ and $b$ work for the desired result.
The solution manual didn't prove this, but I did (for the case of this specific $f(x)$). I am wondering if I did unnecessary work.
For the record, here is the solution manual solution

Let $g(x)=\int_a^x f - \int_x^b f$. Then $g$ is continuous and
$g(a)=-\int_a^b f$ and $g(b)=\int_a^b f$; so $g(a)$ and $g(b)$ have
different signs and consequently $g(x)=0$ for some $x$ in $[a,b]$,
unless $g(a)=0$, in which case we can choose $x=a$.
For the function $f$ shown below, only $x=a$ or $x=b$ will work; $f$
has been chosen so that $\int_a^c f=-\int_c^b f$.


 A: Your proof is good, and no, you didn’t prove that there isn’t an $x$ with this property. You don’t have enough information in the question to say any more or less.
However you can succinctly express the condition on $x$, when $\int_a^b f(t)\,\mathrm{d}t=0$, as: “$x$ is a root of $F$”. Then by continuity, $F$ has no other roots if it has constant sign (for example). Whether or not there is another root is your only criterion for the existence of such $ x\in(a,b)$. This will sometimes be the case and sometimes it won’t; $f(x)=\sin x$ on $[0,2\pi]$ is an example where there isn’t $x\in(a,b)$ but $f(x)=\sin(x)$ on $[0,4\pi]$ is an example where $x=2\pi\in(a,b)$ suffices.
A: Let $f(x)=2x$ on $[-1,1].$
Then
$$\int\limits_{-1}^x 2t\,dt=x^2-1,\qquad \int\limits_{x}^1 2t\,dt=1-x^2$$ The equality holds only for $x-\pm 1.$
The same holds for any odd function $f(x)$ on the interval $[-a,a]$ such that $f(x)$ has constant sign on $[-a,0].$ In particular your example in  $\color{green}{\textrm{green}}$ and $\color{blue}{\textrm{blue}}$ is fine as the function is odd with respect to $(a+b)/2.$  Also the second example is fine.
