2
$\begingroup$

So I'm refreshing on integration and venture across a property of definite integrals I've been taking for granted for quite some time:

\begin{equation}\int_{a}^{b} f(x) \mathrm{d}x =-\int_{b}^{a} f(x) \mathrm{d}x \end{equation}

Stewart's Calculus claims this property is owed to $\Delta x= \dfrac{b-a}{n}$, where $\Delta x$ becomes negative when you flip $a$ and $b$. If this is true, then it would be true for the (I know it's not the exact) definition of a definite integral:

$$\int_{a}^{b} f(x) \mathrm{d}x =\lim_{n \to \infty} \sum_{i=1}^{n}f(x_i^*)\Delta x$$

So I just did a quick verification of a simple definite integral of $x^2$ from $0$ to $1$. Using the Riemann sum definition I got $1/3$ and switching the limits of integration I got $-1/3$ as expected. However, I noticed that if you take $\Delta x$ to always be positive, in the sense that it's a width of a rectangle and couldn't possibly be negative, then the answer either way would be $1/3$. Which, in conclusion, brings me back to my question: wouldn't it be simpler to take $\Delta x$ as a distance or width (always positive) and eliminate this property of integrals? Leaving us with:

$$\int_{a}^{b} f(x) \mathrm{d}x =\int_{b}^{a} f(x) \mathrm{d}x$$

To me this makes more sense, because the area under a curve should be maintained when calculated from any direction.

$\endgroup$
  • $\begingroup$ I did notice that for this to work with the normal way we do definite integrals we would have to take the absolute value of the difference of the antiderivative evaluated at those limits of integration. $\endgroup$ – Zduff Dec 16 '15 at 0:54
  • 2
    $\begingroup$ It's actually far better to have the signed value. Then the integral behaves well with respect to changes of variables, and there are other perks. $\endgroup$ – Matt Samuel Dec 16 '15 at 0:57
4
$\begingroup$

One thing that's definitely simpler with the standard notation is this: The formula $$\int_a^bf(x)dx+\int_b^cf(x)dx=\int_a^cf(x)dx$$is valid with no restrictions on $a,b,c$. With your revision it would be true only if $a\le b\le c$ or $c\le b\le a$. I've always assumed that this was the reason for the convention being the way it is.

$\endgroup$
  • $\begingroup$ Not sure what you mean here. I thought that for this formula to work $b\in [a,c]$ $\endgroup$ – Zduff Dec 16 '15 at 1:24
  • 1
    $\begingroup$ That would be so with the OP's revision. But with the standard notation, $\int_0^2 + \int_2^1=\int_0^1$. (Because that's the same as $\int_0^2=\int_0^1+\int_1^2$.) $\endgroup$ – David C. Ullrich Dec 16 '15 at 1:55
3
$\begingroup$

In some contexts, such as areas, taking absolute value of the integral makes perfect sense.

In others, such as computation of work, or electric charge, or any other signed quantity, it does not.

You can always apply absolute value when needed. You can't "un-apply" it if it is built into the definition of the integral.

$\endgroup$
3
$\begingroup$

$\int_a^b f$ is a misleading notation. What is really defined is the integral over the interval, the set. There is no "orientation" involved. Think of this as $\int_{[a,b]} f$ instead.

Now, we define the symbols $\int_a^b f:=\int_{[a,b]} f$ and $\int_b^af:=- \int_{[a,b]} f$. Why? Because this is convenient. This makes a lot of formulas easier to write, and a lot of cases can be condensed in a single statement (for instance, as commented, the change of variables formula).

$\endgroup$
  • $\begingroup$ Can the downvoter explain his reasoning? $\endgroup$ – Aloizio Macedo Dec 16 '15 at 1:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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