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When dealing with derivatives, "with respect to $x$" means we are observing how a small change in $x$ (the input) affects a change in $y$ (the output).

I found this conceptualization very helpful and it made other derivative related concepts feel more intuitive.

I'm wondering if there is a similar conceptualization of what "with respect to $x$" means when integrating. In particular, how does the input, $x$, affect or relate to the output, $y$, when integrating?

I should say that I'm familiar with the geometric conceptualization of an integral, namely the Riemann sum, and that integrating with respect to $x$ means using the $x$-axis as the lower bound (or base) of the curve when calculating area. Alternatively, one can integrate with respect to $y$ and then the $y$-axis is used as a bound instead. However, it is difficult for me to glean from the geometric interpretation what "with respect to $x$" means when integrating.

This question is motivated by using $u$-substitution requires integration with respect to $u$, but there is no $u$ axis to use as a base to find the area with. I'm sure my understanding of this is incorrect, hence why I'm hoping that better understanding what "with respect to __" means when integrating will help me better understand u-substitution and other integration concepts, much like how understanding what "with respect to __" means when differentiating helped me better understand the Chain Rule.

In shot my main question is:

What does "with respect to __" mean when integrating, as in how does the input affect or relate to the output when finding the area under the curve? Is there a conceptualization along similar lines to what "with respect to __" means when differentiating?

Slecker
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  • I am not sure - did my response answer your question(s)? – user2661923 Sep 22 '20 at 03:17
  • @user2661923 Thank you for your answer, but I'm still not sure what "with respect to __" means when integrating other than the geometric interpretation. I was hoping it might have a similar conceptualization to what "with respect to __" means when differentiating - that we are observing how a change in the input affects a change in the output. In other words, I'm wondering if integrating "with respect to __" indicates how the input affects the output when finding the area under the curve. – Slecker Sep 22 '20 at 03:26
  • @user2661923 Also, I realize that I have asked several questions when my purpose was just to provide context and details. I should have a TL;DR or something that highlights my main question. Apologies. – Slecker Sep 22 '20 at 03:28
  • First of all, I am unfamiliar with your TL;DR notation. I will take a crack at it; tell me if this works. Considering $f'(x)$, changing $x=x_1$ to $x=x_2$ (for example) means that you are saying : I continue to be interested in the slope of $f(x)$ at various points, but I wish to examine the slope at $x_2$, rather than $x_1.$ If you start with $\int_a^b f(x) dx,$ and change this to $\int_a^c f(x)dx$ (for example) you are saying : I want to continue to examine the area under the curve $y = f(x)$, but I which to alter the region that I am examining from $a \to b$ to $a \to c$. see next comment – user2661923 Sep 22 '20 at 03:38
  • Does this answer your question? – user2661923 Sep 22 '20 at 03:38
  • @user2661923 Ah sorry, TL;DR is a short-hand for "too long; don't read," but I've changed it to a recap of my main question instead. – Slecker Sep 22 '20 at 03:41
  • If I understand correctly, when your are examining $f'(x)$, making a small change to $(x + \Delta x)$ means that you are now examining the slope of the curve at $f(x + \Delta x)$. If you are examining $\int_a^b f(x) dx$, and change this to $\int_a^{b + \Delta x} f(x) dx$, you have added a region to the integral whose area is approxmately that of a rectangle of width $(\Delta x)$, and whose height is $f(b)$. Does this answer your question? – user2661923 Sep 22 '20 at 03:46
  • @user2661923 I believe I understand what you are saying, but I don't see how it answers what "with respect to __" means when integrating. Perhaps, I'm mistakenly thinking that there is more to it than there actually is, and all it indicates is which axis is used as the lower bound when finding the area under the curve. I greatly appreciate your efforts to help me understand, however. :). – Slecker Sep 22 '20 at 03:53
  • Did Rivers McForge's response answer your question? – user2661923 Sep 22 '20 at 03:54
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    By the way : some mathSE reviewers might complain that it is unclear what you are asking. Although there are situations where I personally regard such a complaint as reasonable, this is not one of them. I think that the lack of clarity is not your fault, but rather because you are trying to reach for a deep question, focusing on a deep idea. I feel that it is not uncommon to have trouble precisely articulating a question when it involves such depth. – user2661923 Sep 22 '20 at 04:00
  • @user2661923 It's not the answer I was looking for, but it is something I wasn't aware of since all the functions I have integrated thus far have been functions of a single variable like $x$. So I do appreciate his answer for that aspect. – Slecker Sep 22 '20 at 04:00
  • @user2661923 Yes, I wasn't sure how to approach nor succinctly articulate such a question. Hence all the context and details. I do apologize and I do strive for clarity when asking my questions, but in this case, I seem to have failed. – Slecker Sep 22 '20 at 04:05
  • @user2661923 By the way I have upvoted both your and Rivers' answers since both have provided me insight on other aspects of integration. :) – Slecker Sep 22 '20 at 04:07

4 Answers4

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A lot of integral formulas have other variables than $x$ floating around inside them, e.g. $$\int \frac{k dx}{x^2 + a^2} = \frac{k}{a} \arctan \frac{x}{a} + C,$$ and the $dx$ formalism is necessary to specify which of the variables is the dummy variable of integration. We say that the above integral is taken "with respect to $x$" to clarify that it is not being taken with respect to $k$ or to $a$.

Rivers McForge
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The traditional notion is that to evaluate $\int_a^b f(x) dx$, the x axis is the horizontal axis, the $f(x)$ axis or $y$ axis [if you presume that $y = f(x)$] is the vertical axis, and the integral represents the area under the curve in the region bounded by $x=a$, and $x=b$.

When you make a $u$ substitution, this simply means that you have transformed the original integral into something that looks like $\int_c^d g(u) du$, where the u axis is the horizontal axis, the $g(u)$ axis is the vertical axis, and the integral represents the area under the curve in the region bounded by $u = c$ and $u = d.$

By the way, one can construe all of the above as vertical integration, re you are looking for the area under the curve, rather than the area to the left of the curve.

Suppose that you are given $y = f(x)$, and you are asked for the area to the left of the curve in the relevant region. You then have two choices. Compute $g(y) = x$ (if feasible) where $g$ is the inverse function of $f$. Then it would make sense to set up an integral that looks like $\int_e^f g(y) dy$, where $e,f$ represent the corresponding $y$-value bounds for the integral. I would construe this to be integrating horizontally, because you are looking for the area to the left of the curve $y = f(x).$

The alternative and often much easier approach is to attack the problem simply by insisting on computing the area under the curve (i.e. vertical integration). Then, assuming that you have precisely identified the length and width of the pertinent rectangle, and have determined that the area of this rectangle $= R$, then you have that the (area to the left of the curve) + the (area under the curve) = R.

user2661923
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I dislike the idea of defining integration as the area under a curve. You have pointed out some of the reasons in the question.

It is true that you can use an integral to measure the area of a figure that is bounded by a horizontal axis below, vertical lines on the left and right, and the graph of a function above. But this is merely a consequence of how we define integrals and area and graphs of functions. It is not the meaning of an integral.

In my mind, a definite integral means an accumulation of something. It is intimately related to a derivative by the following equation:

$$ \int_a^b \left(\frac{\mathrm dF(x)}{\mathrm dx}\right) \mathrm dx = F(b) - F(a). $$

We have a quantity $F(x)$ that changes as $x$ changes. As you know, the derivative $\dfrac{\mathrm dF(x)}{\mathrm dx}$ describes how quickly some quantity $F(x)$ increases or decreases with respect to a change in the quantity $x$. The integral simply represents that we change $x$ in a continuous way from $x = a$ to $x = b$ and see how much the resulting changes in $F(x)$ accumulate.

The integral means exactly the same thing when written in the more usual way:

$$ \int_a^b f(x)\, \mathrm dx. $$

The function $f(x)$ still tells us how quickly the quantity measured by the interval changes with respect to any change in the variable $x$. That's why it still makes sense to say "with respect to $x$" here.

With a $u$-substitution in the integral, we're still measuring the same accumulation of the same amount of stuff. We've just changed how we measure the rate at which it accumulates. Instead of changing $x$ from $a$ to $b$ we change $u$ from $c$ to $d$, and since $u$ doesn't change at the same rate as $x$ the "stuff" doesn't accumulate at the same rate with respect to $u$ as it does with respect to $x$. The trick to doing a correct $u$-substitution is to end up with something of the form

$$ \int_c^d h(u)\, du $$

where $h(u)$ represents the correct rate of accumulation of "stuff" with respect to $u$ corresponding to the rate of accumulation $f(x)$ with respect to $x$.

This explanation may not make sense at first. I did not really "get" $u$-substitution intuitively until sometime in the second year of undergraduate math education after the first year of calculus.

Another explanation is that $u$-substitution is an "inverse" of the chain rule in the same way that integration is an "inverse" of differentiation. There's an answer to the question "Connection between chain rule, u-substitution and Riemann-Stieltjes integral" that explains this nicely, so I refer you to that.

See my answer to "What's an intuitive explanation for integration?" for more discussion of the meaning of an integral vs. the area under a curve, with some emphasis on applications to physics.

David K
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In general, dx means that the function is integrated with respect to x. $d\alpha$ is with respect ot alpha,$d\beta$ is with respect to beta, and so on.
When looking at an integral that is dx,everything else is a constant. For example, consider the integral $$\int_a^b f(x)g(\alpha)dx$$.Since our integrand is with respect to x, g($\alpha$) is purely a contant! It doesnt change! $$g(\alpha)\int_a^b f(x)dx$$This is because we can treat g$(\alpha)$as a constant, and remove it from the integral.

TLDR; with respect to something means that everything else is a constant, and can be treated as such when solving.

Greatness
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