46
$\begingroup$

Let us define a very simple integral:

  • $f(x) = \int_{a}^{b}{x}$

for $a,b\ge 0$.

Why do we have the identity $\int_{a}^{b}{x} = -\int_{b}^{a}{x}$?

I drew the graphs and thought about it but to me integration, at least in two-dimensions, is just taking the area underneath a curve so why does it matter which direction you take the sum?

| cite | improve this question | |
$\endgroup$
  • 6
    $\begingroup$ Does it perhaps change the sign of $\Delta x$ when viewing the integral as a Riemann sum? $\endgroup$ – OnceUponACrinoid Jun 8 '15 at 2:59
  • 1
    $\begingroup$ But, I know the Riemann sum is not the only method of integration. I just checked and the $\Delta x$ changes when you also look at the Darboux integral. But, in general, regardless of definition, is there an intuitive\graphical meaning as to why it is the opposite sign? $\endgroup$ – David South Jun 8 '15 at 3:08
  • 1
    $\begingroup$ This is related. $\endgroup$ – Joel Bosveld Jun 8 '15 at 7:15
  • 1
    $\begingroup$ Generally, going to the right means "positive" and going to the left means "negative". So, measuring the area from left-to-right gives us a positive signed area, and measuring the area from right-to-left gives us a negative signed area. ("Signed area" just means "area where we care about positives and negatives". Using signed area is convenient, because lots of our formulae end up easier. For example, you already know that $\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx$; if we didn't use signed area, this wouldn't work for $c=a$.) $\endgroup$ – Akiva Weinberger Jun 9 '15 at 3:23
  • 3
    $\begingroup$ Why does "travelling North" change to "travelling South" when you change the sign of your velocity? Basically it's the same concept. $\endgroup$ – Carl Witthoft Jun 9 '15 at 12:50
95
$\begingroup$

Here's another intuitive justification. The obvious graphical intuition says that when $a \leq b \leq c$, then $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$. If we want this formula to hold for arbitrary $a,b,c$, then we should be able to take $a=c$, so that $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx$. But $\int_a^a f(x) dx = 0$, so if we want this formula to hold, we need $\int_a^b f(x) dx = -\int_b^a f(x) dx $.

| cite | improve this answer | |
$\endgroup$
  • 11
    $\begingroup$ +1 for this. Fundamental theorem of calculus is actually somewhat nontrivial, and one does not need it to justify the definition $\int_a^b f(x) dx = -\int_b^a f(x) dx $ for Riemann-integrals. $\endgroup$ – JiK Jun 8 '15 at 7:07
  • 5
    $\begingroup$ This is not just an intuitive justification, but the only justification for this definition. +1 for bringing it up. $\endgroup$ – Paramanand Singh Jun 8 '15 at 8:12
  • 10
    $\begingroup$ I agree: first of all, the fact that $\int_b^af(x)\,dx=-\int_a^bf(x)\,dx$ is by definition, at least in the most widespread introductions to integrals. Of course it's a good definition, that agrees well with the important fact that $\int_a^b+\int_b^c=\int_a^c$ for $a<b<c$. Using the FTC for justifying this definition is, in my opinion, wrong, because the proof of the FTC usually depends on this property (unless one is strict in using positive increments and negative decrements and proves the existence of the derivative with the limits from the left and from the right). $\endgroup$ – egreg Jun 8 '15 at 9:21
  • $\begingroup$ Does this hold even if one of the bounds is infinite? $\endgroup$ – david1201 May 21 at 17:01
41
$\begingroup$

$$\int_{a}^{b} f(x)dx=F(b)-F(a)=-(F(a)-F(b))=-\int_{b}^{a}f(x)dx$$ by the fundamental theorem of calculus.

Or graphically, $$-\int_{b}^{a}f(x)dx=\int_{b}^{a}-f(x)dx $$ and $$-f(x)$$ has the same area as $$f(x)$$ but under the x axis, so the signed area changes

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ this makes sense, thank you! $\endgroup$ – David South Jun 8 '15 at 3:09
  • 10
    $\begingroup$ This is a convention, used to make the Fundamental Theorem of Calculus valid even when the endpoints are backward. $\endgroup$ – GEdgar Jun 8 '15 at 17:57
  • $\begingroup$ What is the second part of this answer (starting with "Or graphically,") intended to show? $\endgroup$ – user253751 Jun 9 '15 at 3:22
  • $\begingroup$ i would never post such trivial answer even if i can earn 1k reputation by it (your answer is fine , i just blame the logic followed here) $\endgroup$ – Abdou Abdou Jun 12 '15 at 14:40
29
$\begingroup$

We want $$\int_a^b +\int_b^c =\int_a^c.$$ now take $c=a$

| cite | improve this answer | |
$\endgroup$
20
$\begingroup$

It seems to me that the notion that integration "is just taking the area underneath a curve" is what leads to confusion here. See this other question for another example of how this notion got someone into trouble and made it difficult for them to use calculus.

Integration is really the measurement of the accumulated effect of something occurring at a particular rate with respect to something else. We could use it, for example, to figure out how much water is in a reservoir if we know the net rate of flow of water into the reservoir at each time during an interval of time. When the net rate of flow "in" is positive during an interval, the amount of water increases from the start to the end of that interval. When the net rate of flow is negative (it is actually flowing "out"), the amount of water decreases from the start to the end of that interval.

If you integrate "backward" (from the end of the time interval to the beginning, instead of from the beginning to the end), it is like playing back a video in reverse: whatever happened during that time interval is undone. The result is exactly opposite what happens when you integrate "forward."

An integral happens to coincide with "area under the curve" when the curve is above the $x$ axis and you integrate from left to right. Imagine you hold a straightedge parallel to the $y$ axis and move the straightedge left or right. When the curve is above the $x$ axis, the height of the curve at any point represents the rate at which the area under the curve to the left of a the straightedge will increase as you move the straightedge past that point to the right. If $a<b$ and you sweep the straightedge over the curve from $x=a$ to $x=b$ (from left to right), area under the curve is added to the left of the straightedge; if you sweep from $x=b$ to $x=a$ (from right to left), area is taken away. The result of one complete round-trip of the straightedge is exactly zero, that is, you end up with exactly what you started with.

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

One less obvious answer is that we really want integration to be the inverse of differentiation in some fairly strong sense. So whatever definition we choose had better satisfy some version of fundamental theorem.

It turns out that in order for that to work, we need integrals to have measure area in a "signed way." Mattice's answer demonstrates this.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Apply the substitution,

$$y = (a+b) - x \iff dy = -dx.$$

The integral becomes, \begin{align*} \int_a^b x dx & = -\int_b^a u du. \end{align*}

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

There are actually several distinct (but related) notions of integration.

It seems to me that your conception of the integral takes place over a set, not a parameterized interval. As such it might be more appropriate to write:

$$ \int_{[a,b]} f(x)dx $$

rather than $$\int_a^b f(x)dx$$ for the "signed area" model of integration you have in mind.

Later on in your studies, this signed area perspective will be formalized as the integral of a function with respect to a measure.

The integral we generally teach in a first calculus course actually depends on a parameterization of the interval we are integrating over, and perhaps most naturally generalizes to the concept of integrating a differential form.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

This certainly isn't a proof. But it does seem to make sense to me. You need to look at it as a line integral.

To compute $\int_0^3 f(x) dx$, we can start by choosing breakpoints $\{0, 1, 2, 3\}$ for the interval $[0, 3].$ It's important to notice that the sequence goes from $0$ to $3$. Then there exists $\{\bar x_0, \bar x_1, \bar x_2\}$ such that $$\int_0^3 f(x) dx = f(\bar x_0)(1-0) + f(\bar x_1)(2-1) + f(\bar x_2)(3-2)$$

To compute $\int_3^0 f(x) dx$, we start by choosing breakpoints $\{3, 2, 1, 0\}$ for the interval $[0, 3],$ where the sequence now goes from $3$ down to $0$. Then there exists $\{\bar x_0, \bar x_1, \bar x_2\}$ such that $$\int_3^0 f(x) dx = f(\bar x_2)(2-3) + f(\bar x_1)(1-2) + f(\bar x_0)(0-1)$$

Then clearly $\int_0^3 f(x) dx = -\int_3^0 f(x) dx$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The change in $x$, $dx$ becomes negative. So the integral sums over negative quantities.

This is useful for when both coordinates of a curve are functions, and you find the area under the curve by $\int_{t_1}^{t_2}y(t) \, dx(t)$. When the $x(t)$ path travels in the negative direction, $dx(t)$ is negative and so the integral decreases.

| cite | improve this answer | |
$\endgroup$

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