Definite integrals with equal limits

Question

As a property of definite integrals, we teach that definite integrals are zero if the lower and upper limits are the same (Wolfram mathworld says this too). Is this valid in general?

In the case of integrating $\dfrac1{4+\sin^2x}$ from $0$ to $π$ using the substitution $t = \tan x,$ we get both new limits $0,$ so the integral should be $0$, but if you use the symmetry of the graph and describe the integral in terms of area, you get a positive value. How would explain this to students? Do we need to say that there are exceptions?

Answer 1

How would explain this to students?

Explain that checking that conditions are satisfied is part of the process of applying theorems/laws.

(Admittedly, though, since integration change of variable is seldom rigorously taught or even presented as a theorem, it is not a great example to demonstrate carefully reading theorems.)

1. In the OP's example, writing the substitution $t=\tan x$ as $t=g(x),$ the variable-change theorem requires that $g'$ is integrable on the original integration domain $[0,\pi];$ since this isn't satisfied, the substitution isn't applicable.

2. A related fake trick is $$\int_a^b f(x)\,\mathrm dx\ne0 \\\text{and}\\u=\color{red}{g(x)} \text{ such that }g’\text{ is integrable on }[a,b]\text{ and }g(a)=k=g(b)\\\implies\\\int_a^b f(x)\,\mathrm dx=\int_k^k\color\red{\text{[this integrand doesn't matter]}}\,\mathrm du=0.$$ Here, in neglecting to attempt to determine the new integrand (doing this is another aspect of verifying that this theorem is applicable), the student is denied the discovery that the given integrand $f$ is not actually compatible with this particular choice of $g.$

Answer 2

ryang has explained the issue, but I think it might be helpful to students to frame the issue in a more positive light--instead of "you can't use this substitution here", say "here's what you need to do to use this substitution." The problem is that $\tan x$ is not differentiable on $[0,\pi]$ because it's not even defined at $x=\frac{\pi}{2}$. So in order to use this substitution you have to find a way to rewrite the integral to avoid the troublesome point. In this example we can write \begin{align} \int_0^\pi\frac{1}{4+\sin^2x}dx&=\int_0^{\pi/2}\frac{1}{4+\sin^2x}dx+\int_{\pi/2}^\pi\frac{1}{4+\sin^2x}dx\\ &=\lim_{r\to\pi/2^-}\int_0^r\frac{1}{4+\sin^2x}dx+\lim_{r\to\pi/2^+}\int_r^\pi\frac{1}{4+\sin^2x}dx. \end{align} The purpose of introducing the limits is to keep $\frac{\pi}{2}$ out of the interval. Now your substitution works fine and you get $$ \lim_{s\to\infty}\int_0^s\frac{1}{5t^2+4}dt+\lim_{s\to-\infty}\int_s^0\frac{1}{5t^2+4}dt. $$ In this example since the integrand is an even function you can just double the first integral.

By explaining things this way, you don't have throw away a promising substitution and, by going through the process of choosing intervals carefully, you reinforce the idea of checking the needed conditions. My feeling is that by the time students are learning these kinds of substitutions they should already be strongly conditioned to recoil against any attempt to do a definite integral using the fundamental theorem of calculus over an interval where the integrand is not defined at every point. For an example of what can go badly wrong, consider how if you naively apply the fundamental theorem to $$ \int_{-1}^1\frac{1}{x^2}dx $$ you get obvious nonsense (a finite negative answer where the integrand is positive and, properly treated, the integral is infinite).

Answer 3

Your question is similar to the issue raised in this post:

Should $\varphi$ be monotone in the integration by substitution?

But in your situation, there is a more critical issue in that your new variable $g(x)=\tan x$ is not defined at $x=\pi/2$.