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I know how to do definite integrals and u-substitution. I am not clear on the notation for how to represent the lower and upper limits when switching variables in terms of $x$ to $u$ as shown in the examples below. I am looking for the exact notation for this.

Consider $\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)dx$. Clearly, $u=\tan(x)\implies du=\sec^2(x)dx$.

Would it be correct to say $\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)dx=\int_{x=-\pi/4}^{0}udu$?

Also, is the notation correct that when one switches the integral from $dx$ to $du$ it is assumed that the limits of integration take on values from $u(a)$ to $u(b)$, so that

$\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)dx=\int_{-\pi/4}^{0}u\cdot sec^2(x)dx=\int_{-1}^{0}udu$?

This notation would make sense in my mind but I am not sure what is technically correct. Any help with this would be much appreciated!

William Garske
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  • Presumably, you have learned to perform a change of variables (or "$u$-substitution") out of some text. That text should include a theorem which explains not only how to perform such a change of variables, but also good notation for it, as well as some kind of proof. Have you looked at your text? – Xander Henderson Dec 08 '23 at 13:29
  • They completely change the terms of $x$ to terms of $u$ in one move which would make this notation obvious. That being said, I want to be more exact with this. I know that the AP Board for the AP Calc AB exam would allow it to be replaced with an indefinite integral and students would still get it right as long as it gets replaced with the correct values at the end. I am just trying to get more precise to show the steps missing in the middle. – William Garske Dec 08 '23 at 13:33
  • What "steps missing in the middle"? Again, what does the theorem in the text actually say? – Xander Henderson Dec 08 '23 at 13:50
  • They show $\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)dx=\int_{-1}^0udu$ which would avoid ambiguity. However, I am wanting to show more steps than this than just jump to the end result. My textbook says this holds true by the Fundamental Theorem of Calculus and by substitution. I don't think this is a theorem question but rather a question about notating how to change the $u-$interval for the $x-$interval. – William Garske Dec 08 '23 at 13:56

2 Answers2

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Consider $$\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)\,\mathrm{d}x.$$ Take $u=\tan(x)$ so that $\mathrm{d}u=\sec^2(x)\,\mathrm{d}x$.

Would it be correct to say $$\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x)\,\mathrm{d}x=\int_{x=-\pi/4}^{0}u\,\mathrm{d}u?$$

No.

When one writes $$\int_{a}^{b} f(t)\,\mathrm{d}t,$$ $a$ and $b$ are the minimum and maximum values of the variable of integration (in the case of the integral above, that variable is $t$). If one wanted to be very specific about this, one could write $$\int_{t=a}^{t=b} f(t)\,\mathrm{d}t.$$

The notation $$\int_{x=-\pi/4}^{0}u\,\mathrm{d}u$$ is very unclear. What is $x$? Where did it come from? What is it supposed to mean? It is confusing. If I saw that notation, I would assume that the integral came from a problem in which the lower bound of integration is a variable, and it is being evaluated for a specific value of that variable. Hence I would evaluate the integral as $$\int_{x=-\pi/4}^{0}u\,\mathrm{d}u = \frac{1}{2} u^2 \Bigr|_{u=-\pi/4}^{0} = \frac{1}{2} \left( -\frac{\pi}{4}\right)^2 = \frac{1}{32} \pi^2. $$

The relevant theorem, as presented in Thomas' Calculus (early transcendentals, 13th ed), is

Theorem: If $g'$ is continuous on the interval $[a,b]$ and $f$ is continuous on the range of $g(x) = u$, then $$ \int_{a}^{b} f(g(x))\cdot g'(x)\,\mathrm{d}x = \int_{g(a)}^{g(b)} f(u)\,\mathrm{d}u. $$

In the given problem, take $f(x) = x$ (or, equivalently, $f(u)=u$) and $u = g(x) = \tan(x)$. Observe that $g'(x)=\sec(x)^2$ is continuous on the interval $[-\pi/4,0]$. $g$ is monotonic on that interval, hence the image of that interval is $$ g([-\pi/4,0]) = [g(-\pi/4), g(0)] = [-1,0]. $$ Moreover, $f$ is continuous on on this interval, hence the hypotheses of the theorem are satisfied. Doing a little bit of scratch work, note that $$ f(g(x)) = \tan(x) \qquad\text{and}\qquad g'(x) = \sec(x)^2. $$ The theorem can then be applied to conclude that $$ \int_{-\pi/4}^{0} \tan(x)\sec(x)^2\,\mathrm{d}x = \int_{-1}^{0} u\,\mathrm{d}u.$$

Again, the theorem asserts that if you have an integral with certain properties, then you may replace that integral with another integral. There are no "intermediate steps"—the theorem is a "black box": you feed it an integral of a given form, and it spits out another integral which evaluates to the same number.

Xander Henderson
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  • Would it be correct to write $\int_{-\pi/4}^0 \tan(x) \cdot \sec^2(x),\mathrm{d}x=\int_{x=-\pi/4}^{x=0}u,\mathrm{d}u$? – William Garske Dec 08 '23 at 14:36
  • @WilliamGarske I wouldn't. As I said above, what is $x$? That is super confusing. Again, the theorem is a black box. You feed it an input, it spits out an output. Why do you want to include unnecessary and confusing "intermediate steps"? – Xander Henderson Dec 08 '23 at 15:05
  • @WilliamGarske It is done via application of the theorem. There is nothing to notate. – Xander Henderson Dec 08 '23 at 15:07
  • Because sometimes the expressions are a lot longer. It gets to be easier to substitute $u$ in not all at once. – William Garske Dec 08 '23 at 15:07
  • Then you do the work ahead of time, when you are declaring the meaning of the variables and functions. Note the exposition that I included before writing down the final result. – Xander Henderson Dec 08 '23 at 15:09
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Let $f(x)$ be a real-valued function which is bounded on $[a,b]$, and let $g(x)$ be a monotonically increasing differentiable real-valued function. Suppose also $f(x)$ and $g’(x)$ are Riemann integrable. Then the Riemann integral $\int_a^b f(x)g’(x)dx$ is equal to the Riemann-Stieltjes integral, denoted as $\int_a^b f(x)dg(x)$.

The precise definition of Riemann-Stieltjes integral, as well as a proof of the above property, may be found in any relevant text. For the purpose of this answer, the property will suffice.

Corresponding to your example, we have $a=-\pi/4,b=0, f(x)=\tan(x)\sec^2(x),g(x)=x$ initially. But we may also take $f(x)=g(x)=\tan(x)$, because $g’(x)=\sec^2(x)$. Then we may replace $\tan(x)$ by $u$. Note that this replacement is not your standard u-substition.

That is, $\int_{-\pi/4}^0 \tan(x)\sec^2(x)dx=\int_{-\pi/4}^0 \tan(x)\text{d}\tan(x)=\int_{-1}^0 u du$.

In this sense, the middle integral might be written as $\int_{-\pi/4}^0 \tan(x) dg(x)$, provided that you have declared $g(x)=\tan(x)$ beforehand. However, simply writing $dg$ instead of $dg(x)$ would still cause confusion, and should not be done.

Divide1918
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  • The part that I am finding confusing is that we can't simply substitute a variable in the integrand without confusing the variables of integration. – William Garske Dec 08 '23 at 17:01
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    @WilliamGarske This is why I wrote that it’s important one writes $dg(x)$ and not merely $dg$ – Divide1918 Dec 08 '23 at 17:03
  • That makes sense to me. Thank you for clarifying. – William Garske Dec 08 '23 at 17:31
  • I really appreciate you helping me. That's such an ugly notation though. Because then you are going to want to call $g(x)$ everything in the integrand. Have you ever seen it written like $\int_{-\pi/4}^0 u du(x)$ and then go to the next step just to avoid writing $u(x)$ all the time in the integrand? – William Garske Dec 08 '23 at 17:58
  • @WilliamGarske $g(x)$ is NOT “everything in the integrand”, please read my answer more carefully again. And I never wrote $\int_{\pi/4}^0 udu(x)$, the u in my answer is not a function of x, but a variable just like x itself. – Divide1918 Dec 09 '23 at 04:18