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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?

David South
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    Does it perhaps change the sign of $\Delta x$ when viewing the integral as a Riemann sum? – OnceUponACrinoid Jun 08 '15 at 02:59
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    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? – David South Jun 08 '15 at 03:08
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    This is related. – Joel Bosveld Jun 08 '15 at 07:15
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    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 $c=a$.) – Akiva Weinberger Jun 09 '15 at 03:23
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    Why does "travelling North" change to "travelling South" when you change the sign of your velocity? Basically it's the same concept. – Carl Witthoft Jun 09 '15 at 12:50
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    @OnceUponACrinoid : Can anyone producea definition of the Riemann sum that is defined when the endpoints are in the "wrong order"? All I find is the strict requirement that the partition have the first endpoint ($\int_\text{this one}$) be the first interval boundary, then the sequence increases strictly monotonically, and the last endpoint agrees with the last interval boundary ($\int^\text{this one}$). All I can find is $\int_1^0 1 ,\mathrm{d}x$ is undefined due to nonexistence of a valid partition. – Eric Towers Apr 07 '16 at 18:22

10 Answers10

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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 $.

Gregory J. Puleo
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    +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. – JiK Jun 08 '15 at 07:07
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    This is not just an intuitive justification, but the only justification for this definition. +1 for bringing it up. – Paramanand Singh Jun 08 '15 at 08:12
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    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). – egreg Jun 08 '15 at 09:21
  • Does this hold even if one of the bounds is infinite? – david1201 May 21 '20 at 17:01
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    If $a=c$ that means $a=b$ because $a \leq b \leq c$. Then $\int_a^b f(x) dx = \int_b^a f(x) dx = 0$ meaning that the proof does not really mean anything? – Lanet Mar 26 '21 at 21:06
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    @Lanet Gregory's argument is not a proof. He is just saying that if we want the formula $\int_a^b f + \int_b^c f = \int_a^c f$ to make sense for arbitrary $a$, $b$, and $c$, we have no choice but to define $\int_b^a f= -\int_a^b f$. – Alexander Christiansson Jul 09 '21 at 19:47
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$$\int_{a}^{b} f(x)dx=F(b)-F(a)=-(F(a)-F(b))=-\int_{b}^{a}f(x)dx$$ 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

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

Thomas Andrews
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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.

Community
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David K
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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.

Zach Stone
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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 $\int_0^3 f(x) dx = -\int_3^0 f(x) dx$

Steven Alexis Gregory
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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*}

Jester Tran
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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.

Steven Gubkin
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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.

wlad
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There is nothing to prove here. It is just a definition (more precisely, just a notaion). Note that for $b \geq a$, from the point of view of Lebesgue integration, the value of the integral only depends on the domain $[a,b]$ (and does not depend on the fact that if we consider the function from $a$ to $b$ or $b$ to $a$) and this value is written as $\int _a^b f$.

The point is that if we make such a definition (or notation), then we can write the calculations involving the change of variable formula for integration in 1D more conveniently. More precisely, if we have such a definition, then we do not need to consider separately the two cases, when the change of variable function is monotonically increasing or decreasing.

abir
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