Understanding integrals double meaning I would like to understand why the integrals represent 2 things:


*

*If we only do an integral without bounds on $f(x)$, the integral gives us the function $F(x)$ that, when differentiated, gives us the original function $f(x)$.

*If we do an integral with bounds $[a,b]$ on $f(x)$, it gives us the area under $f(x)$ by evaluating the difference $F(b) - F(a)$.
Why does $F(x)$ represent the area under $f(x)$ as well as the first point?
 A: In fact, the integrals in 1 and 2 are rather similar symbols for rather different things:


*

*This "integral without bounds" does not define a function, because there are many functions whose derivatives are equal to a given function. So at this point there is no new concept, but just a new symbol for the concept of derivative: $F(x)=\int f(x) dx$ must only be understood as a synonym for $F'(x)=f(x)$.

*There is the problem (in fact much older than derivatives) of computing areas. To formalize it, we represent an area as $\int_a^b f(x) dx$ and define it as the limit of the sum of the small rectangular areas obtained by partitioning the enclosed area in infinitely-many portions.
Then, in order to solve problem 2, we define a function by arbitrarily establishing a left-hand bound of the area and moving the right-hand bound:
$$F(x)=\int_a^x f(t) dt$$
But we have not done anything yet, this is just a tool to rephrase the problem of computing areas as computing values of $F(x)$.
Now, the revolution (by Newton and Leibnitz): we cannot say anything about $F(x)$ itself, but we do know how it increases. The rate of increase of $F(x)$ at a point $x$ is precisely $f(x)$. (The graphical intuition is that any time we move the right-hand bound, we are adding a thin rectangle of height $f(x)$). So this is the impressive achievement known as Fundamental Theorem of Calculus: relating the concepts of derivative and area by the formula:
$$\frac{d}{dx} \int_a^x f(t) dt=f(x)$$
So, the practical consequence is that if you are able to find some function $F(x)$ with $F'(x)=f(x)$, then you know that there is a constant $C$ such that:
$$\int_a^x f(t) dt=F(x)+C$$
Now, substitute at $a$ to get:
$$C=-F(a)$$
and substitute at $b$ to get:
$$\int_a^b f(x) dx=F(b)+C=F(b)-F(a)$$
A: In the first case, to reiterate the admonishing of calculus teachers everywhere, don't forget the $C$! In other words, if $F$ is a differentiable function such that $F'=f$, then so is $F'+C$ for any $C\in\mathbb R$. We would write this
$$\int f(x)\ \mathsf dx = F(x) +C. $$
So really, the indefinite integral maps functions to equivalence classes of functions.
In the second case, this is only true if $f$ is a continuous function.
