Suppose $f$ is defined as follows: $$\begin{align}f : [4, 12] &\longrightarrow \R\\ t &\longmapsto \begin{cases} 3 & \text{ if } 5 \lt t \leq 10 \\ 1 & \text{ otherwise.}\end{cases}\end{align}$$ Show that $f$ does not have an antiderivative.
(Hint: There are several ways of solving this problem. One may take a direct approach: first, show that if $f$ does have an antiderivative, then it must be a piecewise linear function with different slopes on the different pieces; but there is no way such a function could be differentiable where the slope changes. In fact, it is possible to prove a much more general statement: any function with a jump discontinuity has no antiderivative.)
Suppose that $f : [a, b]\to\R$ is continuous, and that $F : [a,b] \to \R$ is continuous, differentiable on $(a,b)$, and $F'(z) = f(z)$ for $z \in (a,b)$. Use the second fundamental theorem to prove the first fundamental theorem by showing that $$\frac{d}{dt}\paren{F(t) - \int_a^t f(x)\,dx} = 0.$$
First, show the identity above. Then, use this to prove the first fundamental theorem: if $F : [a, b] \to \R$ is continuous, differentiable on $(a, b)$, and $F'$ is continuous and bounded on $(a, b)$ then $$\int_a^b F'(x)\,dx = F(b) - F(a).$$
A set $S \subset \R$ is said to be null if for any $\epsilon \gt 0$ there is a sequence of open intervales $\paren{(a_n, b_n)}_n$ such that their union contains $S$ and their combined length is less than $\epsilon$: $$S\subseteq \bigcup_{n=1}^\infty (a_n, b_n) \qquad\qquad\text{ and }\qquad\qquad \sum_{n=1}^\infty b_n-a_n \lt \epsilon.$$
As it turns out, there are null sets which are not countable; the Cantor set is a famous example. Somewhat more surprisingly, the function which is $1$ on the Cantor set and $0$ on its complement is discontinuous precisely on the Cantor set, and therefore integrable by this problem. On the last assignment's bonus problem we showed that there was an integrable function with a countable set of discontinuities; now there is even one with an uncountable set of discontinuities.