8
$\begingroup$

This is a dumb question and I don't really know how to word it. When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0). I can easily see how the derivative of an integral is given by the function value, but why does the integral start at 0 and not any other number? When I try to imagine the area of some curve starting at negative 1 for example the area under the curve would intuitively to me still be given by the antiderivative. 0 makes sense as a starting point but for some reason I can't visualize it. I'm not sure if that made any sense but if anyone could help me wrap my head around it I'd appreciate it.

$\endgroup$

marked as duplicate by BallBoy, астон вілла олоф мэллбэрг, Claude Leibovici integration May 10 '18 at 6:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ It doesn't matter. It's just convenient. As for your example, $F(x)=\int_{-1}^x f(t)dt$ is another anti-derivative of $f$. $\endgroup$ – Yanko May 9 '18 at 19:09
  • 5
    $\begingroup$ "given by the antiderivative" There is no "the antiderivative". There is only "an antiderivative". $\endgroup$ – Arthur May 9 '18 at 19:13
  • 4
    $\begingroup$ $\int dx/x$ certainly doesn't start from $0$! $\endgroup$ – Chappers May 9 '18 at 19:15
  • 2
    $\begingroup$ Starting from somewhere else amounts to changing the constant of integration. $\endgroup$ – amd May 9 '18 at 20:46
  • 1
    $\begingroup$ @LSpice, I may have been misinterpreting what op meant. I was essentially thinking the confusion was about why an integral over a degenerate interval was zero. $\endgroup$ – jdods May 10 '18 at 3:11
19
$\begingroup$

When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0).

This is not true. In certain situations it may be the case, but not generally. I think the reason this confusion arises is that a common problem given to calculus students is to find the antiderivative of a polynomial, e.g.,

$$\int x^3 +2x \, dx = \frac{1}{4}x^4 + x^2 + C$$

and in this case, if we set $C = 0$ we get

$$\frac{1}{4}x^4 + x^2$$

which is the same as $$\int_0^x u^3 +2u \, du = \frac{1}{4}u^4 + u^2 \big|_{u=0}^{u=x}.$$

This will work whenever the form that the antiderivative $F$ of $f$ you get takes satisfies $F(0) = 0$. But in general, setting $C = 0$ will not get you the integral $\int_0^x f(t) \, dt$. For example, if you take $f(x) = e^x$ then $$\int e^x \, dx = e^x + C$$ but setting $C = 0$ gives you $e^x$, which is not the same as $$\int_0^x e^t \, dt = e^x - 1.$$

Note that "setting "C = 0" in the expression for the antiderivative" is not actually a well-defined operation. Different methods of antidifferentiation can give you different expressions when you set $C = 0$. It is important to remember that there is no single antiderivative, and no canonical way of writing it. $\int 2x \, dx = x^2 + 3 + C$ is just as valid as $\int 2x \, dx = x^2 + C$.

$\endgroup$
  • $\begingroup$ Ahh you're e^x example shows that my assumption was wrong. Thanks for clearing it up. $\endgroup$ – conyare May 9 '18 at 19:39
  • 2
    $\begingroup$ My favourite example is arcsin(x)+C vs -arccos(x)+C, these are equal families, but which function is supposed to be C=0? $\endgroup$ – Serge Seredenko May 9 '18 at 20:27
  • $\begingroup$ @SergeSeredenko, or $\sin(x)^2 + C$ versus $-\cos(x)^2 + C$. $\endgroup$ – LSpice May 10 '18 at 1:03
  • $\begingroup$ Also see this recent question where the difference between two solutions doesn't even look like a constant, but it is: math.stackexchange.com/q/2764985/78887 $\endgroup$ – Euro Micelli May 10 '18 at 3:18
3
$\begingroup$

Integrals don't always start at $0$. Let's start from definite integrals and move to the indefinite integrals you asked about. We know $\int_a^b f(x)dx $ gives the area under the curve between $a$ and $b$. If $f(x)$ has an antiderivative $F(x)$, the Fundamental Theorem of Calculus tells us that $\int_a^b f(x) = F(b) - F(a)$. Thus, if we want the area under the curve from $a$ to $b$, we compute $F(b) - F(a)$. If we want the area under the curve from $0$ to $b$, we compute $F(b) - F(0)$. $F(b) - F(0)$, as a function of $b$, gives us the area from $0$ to $b$.

Now there are a lot of "natural" functions where $F(0) = 0$ (e.g., functions like $x^2$ or $\sin x$), so to $F(b)$ gives the area from $0$ to $b$. But that won't be the case if $F(0) \neq 0$.

The above should make it clear that there's no reason $0$ is special -- if you want the area from $-1$ to $b$ as a function of $b$, just use $F(b) - F(-1)$.

$\endgroup$
  • $\begingroup$ I assumed that the fundamental theorem worked because F(a) gets the area to A from 0 and F(b) gets the area to B from 0 so subtracting will get the desired area in-between them. $\endgroup$ – conyare May 9 '18 at 19:18
  • 1
    $\begingroup$ @conyare No, it actually goes the other way around! It is true that if all you knew were the 0-to-b version of the fundamental theorem, you could recover the a-to-b version. But the subtraction of another value is "fundamental" (pun somewhat intended) to the fundamental theorem. Jeffery Opoku-Mensah hints at a reason why this is true: since $F$ is defined only up to a constant, $F(b)$ doesn't give you a definite answer, only up to a constant. Only once you subtract $F(b)-F(a)$ does the constant cancel and you get a definite area. $\endgroup$ – BallBoy May 9 '18 at 19:21
3
$\begingroup$

Well, integrals (by which I think you mean anti derivatives) don't always start at $0$. Indeed, if a function $f=f(x)$ is continuous in some interval $I=[a,b]$, then it has an anti derivative given by $$\int_c^x{f(t)dt},$$ where $c\in I$. The $0$ is usually chosen for $c$ only as a matter of convenience. A well-known function defined by an antiderivative that "starts" from (note that the notion of starting from for antiderivatives should not be taken too literally since the function is defined even for $x<c\in I$) $1$, not $0$ as usual, is the logarithm function $$\log x=\int_1^x{\frac{1}{t}dt}$$ defined for all $x>0$.

$\endgroup$
0
$\begingroup$

The antiderivative is generally given as the "simplest" form, and often the simplest form has a y value of zero when x is zero. For instance, if the function is constant, then for the antiderivative, you need a line whose slop is equal to that constant value. The simplest way of writing that is y = mx+C. You could write y = mx+5-C, and that would be a valid antiderivative, but that would be needlessly complicated. Since you're looking for simple functions, zero will pop up a lot. But there are cases where the simplest function doesn't go through the origin. For instance, if you're taking the antiderivative of sin(x), the simplest form is cos(x)+C. If you want to "start" at zero, you'd have to do cos(x)-1+C. Similarly, the antiderivative of e^x is generally given as e^x+C.

There are some cases where the same function can have different antiderivatives that look very different, but are actually the same thing. For instance, $\frac{1}{\sqrt{1-x^2}}$ can be integrated as either arcsin(x) or -arccos(x). But these just differ by the constant $\pi /2$.

$\endgroup$
-4
$\begingroup$

It depends on what antiderivative you take. For example, an antiderivative of $2x$ can be $x^2$, $x^2+1$, $x^2+100$, or whatever. By the Fundamental Theorem, it turns out $x^2$ gives the area under the curve starting at zero. But $x^2+1$ would also give the area of the curve starting at $-1$, for example.

$\endgroup$
  • 6
    $\begingroup$ I'm afraid you're confusing derivatives with antiderivatives... $\endgroup$ – zipirovich May 9 '18 at 19:22
  • $\begingroup$ Yea, I wrote this in a confused rush, but the idea is still there. Just switch the x^2 and 2x. $\endgroup$ – Jeffery Opoku-Mensah May 9 '18 at 21:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.