What area does the antiderivative represent? Consider the antiderivative of the function $e^{-x},$ which is $-e^{-x}.$ Evaluating the antiderivative at the value $0$ produces $-1.$
I was taught to conceptualize an antiderivative as an area under a curve, or a sum of progressively smaller approximate sections.
But clearly, $-1$ cannot represent the area under under the $e^{-x}$ curve from $0$ to $0$, $0$ to $\infty$, or $-\infty$ to $0$ when you consider that the function is positive for all values of $x$.
Then what sum or area does the value of the antiderivative of $e^{-x}$ actually represent?
 A: 
Evaluating the antiderivative at the value $0$ produces $-1$. Now, I was taught to conceptualize an antiderivative as an area under a curve


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*For $t>a,$ the signed area (i.e., the area in the positive vertical region minus the area in the negative vertical region) $A(t)$ enclosed by the curve $y=f(x),$ the $x$-axis, and the vertical lines at $a$ and $t$ is denoted by the (definite) integral $$\int_a^tf(x)\,\mathrm dx,$$ whose elongated-S symbol stands for Sum (of areas).


*If $f$ is continuous on $[a,t],$ then the Fundamental Theorem of Calculus says that this integral is computable using the formula $$\int_a^tf(x)\,\mathrm dx=F(t)-F(a),\tag1$$ where $F$ is an antiderivative of $f,$ that is, any of the infinitely many functions whose derivative (slope function) is $f.$ Therefore, if $f$ is continuous on $[a,t],$ then $$\frac{\mathrm d}{\mathrm dt}\int_a^tf(x)\,\mathrm dx=f(t).\tag2$$ Rewriting $(1):$ if $F$ is continuously differentiable on $[a,t],$ then $$\int_a^t\frac{\mathrm d}{\mathrm dx}F(x)\,\mathrm dx=F(t)-F(a).\tag1$$


*The indefinite integral $$\int f(x)\,\mathrm dx$$ represents every antiderivative of $f,$ if any. For example, \begin{align}\int \frac1x\,\mathrm dx &= \begin{cases} \ln|x|+C_1, &x<0;\\   
    \ln|x|+C_2, &x>0.\end{cases}\end{align}
Conceptually, an indefinite integral is not technically the precursor of a definite integral: for example,
for \begin{align}g(x)&= \begin{cases} 2x\sin\frac1{x^3}-\frac3{x^2}\cos\frac1{x^3}, &x\ne0;\\   
    0, &x=0,\end{cases}\end{align} $\displaystyle\int_{-1}^1 g(x)\,\mathrm dx$ does not exist even as \begin{align}\int g(x)\,\mathrm dx&= \begin{cases} x^2\sin\frac1{x^3}&+C, &x\ne0;\\   
    0&+C, &x=0,\end{cases}\end{align} that is, even as $g$ does have antiderivatives on $[-1,1].$


*Since antidifferentiation and indefinite integration literally reverse differentiation, they are pointwise operations.
On the other hand, definite integration is a computation with reference to some interval.
The Fundamental Theorem of Calculus ($(1)$ & $(2)$ above) shows the surprising intimate relationship between definite integration and differentiation.


*An antiderivative per se does not describe area, and $F_λ(0)=-1$ means that a particular function whose slope is given by $f$ passes through $(0,-1).$
Clearly, a single value of an antiderivative is not useful information; the FTC's power comes from making its two evaluations using the same antiderivative of $f:$ $$\int_a^tf(x)\,\mathrm dx=F_λ(t)-F_λ(a)\\\neq F_λ(t)-F_ω(a).$$


*A pertinent comment by B. Sullivan:

I just finished reading Steven Strogatz's new book about calculus, Infinite Powers. In it, he lays out the "Three Central Problems of Calculus":

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*Given a curve, find its slope everywhere. [differentiation]

*Given a curve, find another curve whose slope everywhere is that given curve. [antidifferentiation]

*Given a curve, find the area under it. [(definite) integration]




*The actual area enclosed by $y=f(x),$ the $x$-axis, and
the vertical lines at $a$ and $t$ is $$\int_a^t|f(x)|\,\mathrm dx,$$ which does not generally equal $F(t)-F(a).$
