Generalization of the derivative of accumulation function: 1st fundamental theorem of calculus

Question

The 1st fundamental theorem of calculus: $$ \frac{d}{dx} \left[\int_a^x f(t)\,dt) \right]=f(x),\qquad x \in (a,b). $$ No textbook says that this can be generalized into the scenario where $x$ is less than $a$ and where $f$ is continuous for $x<a$. In fact, $x$ can be anything other than $a$, as long as $f$ is continuous in the regions in question. Why is this not generalized in any textbook?

Answer 1

It is a bold claim that no textbook contains the desired generalization, but I am inclined to agree with you as I have never seen one.

The reason for this "omission" is simple: The (Riemann) integral $\int_a^b f(t)dt$ is defined for functions $f : [a,b] \to \mathbb R$ on an interval $[a,b]$, and here we have $a < b$.

If we critically examine this construction, we see that in most textbooks not even $\int_a^a f(t)dt$ is defined, and this integral is needed in order that $$F(x) = \int_a^x f(t)dt$$ is well-defined for $x = a$. It seems that most authors tacitly assume that $\int_a^a f(t)dt = 0$ which is of course justified. In a few textbooks it is essentially covered by the the general construction of the integral if we agree to consider $[a,a]$ as a (degenerate) interval. An example is "Baby Rudin": A partition of an interval $[a,b]$ is a finite sequence of points $x_i$ such that $a = x_0 \le x_1 \le \ldots \le x_{n-1} \le x_n = b$. Working with this concept we see that $\int_a^a f(t)dt = 0$. However, in most textbooks it is required that $a = x_0 < x_1 < \ldots < x_{n-1} < x_n = b$, hence $\int_a^a f(t)dt$ is not covered by the construction, and then $\int_a^a f(t)dt = 0$ becomes a definition. There are good reasons to do so.

1. For all $v \in (u,w)$ we have $$(*) \quad \int_u^w f(t)dt = \int_u^v f(t)dt + \int_v^w f(t)dt .$$ This generalizes to all $v \in [u,w]$ iff we set $\int_a^a f(t)dt = 0$.

2. The above function $F : (a,b] \to \mathbb R$ is continuous. We have $\lim_{x\to a} F(x) = 0$, thus a continuous extension to $[a,b]$ is obtained iff we set $F(a) = 0$.

We now come to your question. Integrals of the form $\int_c^d f(t)dt$ for $c > d$ are not considered in many textbooks, and thus it is no surprise that your generalization does not occur.

Of course, one usually defines $$(**) \quad \int_c^d f(t)dt = - \int_d^c f(t)dt. $$ This has good reasons. In fact, we want that $(*)$ is satisfied for all $u,v,w$ in the interval on which $f$ lives, and this enforces $(**)$. See for example Why does an integral change signs when flipping the boundaries? It has also a nice intuitive interpretation. If we integrate from $d$ to $c$, then we go in positive direction from $d$ to $c$ and any partition $P = (x_0,x_1,\ldots,x_n)$ of $[d,c]$ produces the factors $d_i =(x_{i+1}-x_i)$ in the Riemann sum. If we do it in the other direction, we get the factors $(x_i - x_{i+1}) = -d_i$. This leads to a change of sign of the whole Riemann sum.

For any integrable function $f : [a,b] \to \mathbb R$ and any $r \in [a,b]$ we can now define $$F_r : [a,b] \to \mathbb R, F_r(x) = \int_r^x f(t)dt .$$

For a continuous $f$ we then we have $F_r'(x) = f(x)$ for all $x \in [a,b]$ (see mathworker21's comment). In fact this is a trivial generalization of the standard theorem which says that $F'_a(x) = f(x)$ for all $x \in [a,b]$. Simply note that $F_r(x) = F_a(x) - F_a(r)$.