Is it proper integral notation to use "x=" in the bounds of a definite integral?

Question

So, several years ago I skimmed through a calculus textbook and some of it stuck, some of it didn't. Now several years later I'm finishing up my second Calculus class at university, and I'm a little confused.

If you'd asked me a year ago to write the general form of a definite integral, this is how I would have written it:

$$\int_{x=a}^{b}f(x)\,dx.$$

Now, the obvious outlier is the presence of the "$x=$" in the lower bound. What confuses me is that I'm sure I've seen that before. And I don't think it was from summation notation. And looking around I can't seem to see any signs of it being used anywhere.

Now, some might say that the "$x=$" is entirely redundant, and it largely is. The only exception I can think of with substitution. Typically speaking you either use the u-substitution in an indefinite integral, then place the bounds on the solution of the indefinite integral to find the value of the definite integral, or you convert the bounds with every substitution. But since the bounds don't come into play until the last step, it makes sense to me that you could use this notation to leave the bounds alone without needing to worry about keeping track of what variable they were tied to. Then as you get to the end, it's up to you whether you would like to convert the bounds to be in terms of your final variable or the resulting equation back in terms of the original variable.

So, is this notation (or one similar to it) ever used in mathematics? If it's not in use, has it ever been or is my memory simply playing tricks on me?

Answer 1

The only place that I've seen this used is, as you mentioned, with substitution, where you want to keep track of which variable the bounds are referring to. For example, to compute the integral $$\int_{0}^{2}\frac{dx}{(1+x)^2}$$ we could use the substitution $u = 1+x$, $du = dx$. This would change the bounds to $1$ and $3$ in $u$. If I wanted to keep track of the bounds throughout the solution (as opposed to solving the indefinite integral and then reintroducing the bounds), I would do one of the following:

Method One: $$\int_{0}^{2}\frac{dx}{(1+x)^2} = \int_{x=0}^{x=2}\frac{du}{u^{2}} = -\frac{1}{u}\bigg|_{x=0}^{x=2} = -\frac{1}{1+x}\bigg|_{0}^{2} = -\frac{1}{3} + 1 = \frac{2}{3}.$$

Method Two: $$\int_{0}^{2}\frac{dx}{(1+x)^2} = \int_{1}^{3}\frac{du}{u^{2}} = -\frac{1}{u}\bigg|_{1}^{3} = -\frac{1}{3} + 1 = \frac{2}{3}.$$

Answer 2

I use that notation in "Calculus from the Ground Up". It is helpful because if you have integrals containing multiple variables, you can be more clear.

Example:

$$ \int_{x = 1, y = 0}^{x = 2, y = 7} 2x\, dx + dy = x^2 + y \biggr|_{x = 1, y = 0}^{x = 2, y = 7} $$ $$ ((2)^2 + 7) - ((1)^2 + 0) = 11 - 1 = 10$$

Answer 3

I often use it for readability in iterated integrals, like this: $$ \int_{x=a}^{b} \left( \int_{y=c}^d f(x,y) \, dy \right) \, dx $$ rather $$ \int_{a}^{b} \left( \int_{c}^d f(x,y) \, dy \right) \, dx . $$ Redundant, yes, but then you don't have to look over at the right end to determine which variable the bounds refer to.