4
$\begingroup$

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?

$\endgroup$
3
  • 12
    $\begingroup$ There is no the antiderivative of $e^{-x}$. Every function of the form $-e^{-x} + C$ is an antiderivative of $e^{-x}$. Secondly, a connection between antiderivatives and areas is given by the fundamental theorem of calculus surely, but strictly speaking, it’s definite integrals that should be thought of as an area, not indefinite integrals. $\endgroup$ Commented Sep 7, 2021 at 18:43
  • 2
    $\begingroup$ For $0 \leq a < b,$ you have that $$~\int_a^b e^{-x}~dx = \left[-e^{-x}\right]_{x=a}^{x=b} = e^{-a} - e^{-b}.$$ $\endgroup$ Commented Sep 7, 2021 at 19:20
  • $\begingroup$ Don’t forget that it would just solve $y’=e^x$. Also see:$y’(x,y)=e^{x+y}\implies y=e^{x+y}+f(y)+c$ $\endgroup$ Commented Oct 21, 2021 at 17:25

1 Answer 1

3
$\begingroup$

Evaluating the antiderivative at the value $0$ produces $-1$. Now, I was taught to conceptualize an antiderivative as an area under a curve

  1. 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).

  2. 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 $$\int_a^t\frac{\mathrm d}{\mathrm dx}F(x)\,\mathrm dx=F(t)-F(a).\tag1$$

  3. 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 $[-1,1].$

  4. Since antidifferentiation and indefinite integration literally reverse differentiation, they don't need to be specified on an interval.

    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.

  5. 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 $f:$ $$\int_a^tf(x)\,\mathrm dx=F_λ(t)-F_λ(a)\\\neq F_λ(t)-F_ω(a).$$

  6. 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":

    1. Given a curve, find its slope everywhere. [differentiation]
    2. Given a curve, find another curve whose slope everywhere is that given curve. [antidifferentiation]
    3. Given a curve, find the area under it. [(definite) integration]
  7. 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 $F(t)-F(a).$

$\endgroup$
1

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.