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?
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$\begingroup$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 $.
$\endgroup$ 6 $\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
$\endgroup$ 4 $\begingroup$We want $$\int_a^b +\int_b^c =\int_a^c.$$ now take $c=a$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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 $\int_0^3 f(x) dx = -\int_3^0 f(x) dx$
$\endgroup$ 1 $\begingroup$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*}
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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