I don't have issues with doing integration problems, but occasionally I see the solution changing the limits of integration whenever a $u$-substitution is done.

I obviously don't have a problem doing this, and I just recently noticed my book doing this under the chapter involving "area of surface of revolution."

My question is did I develop a bad habit by never changing the limits of integration, or is it a best practice to always change limits of integration?

My teacher said on a test, and in general, if we do not change the limits of integration then we should be signifying this by labeling our limits of integration $x=$ lower-limit and $x=$upper-limit.


After further investigation, my confusion is because of the two below equations:

Find the exact area of the surface obtained by rotating the curve about the x-axis $$y=\sqrt{1+4x}, 1\le x\le 5$$

The limits of integration were changed in the solution to this problem.

The given curve is rotated about the y-axis. Find the area of the resulting surface. $$y=x^\frac{1}{3}, 1\le y \le 2$$

The limits of integration were NOT changed in the solution to this problem.

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    $\begingroup$ If you never got a result wrong by not changing the limits of integration, you should gamble, gamble a lot. $\endgroup$ – Git Gud Nov 9 '13 at 21:12
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    $\begingroup$ Your teacher is right, forgetting about changing the limits of integration is a frequent source of error. You usually have two choices: (i) Find an antiderivative in terms of $u$, replace $u$ by what it is in terms of $x$, then use the old limits or (ii) Put the limits in terms of $u$, and never go back to $x$. Do one or the other, don't "mix." In general I prefer (ii). Your teacher's advice, writing $x=a$ or $u=c$ will help you keep track, and will prevent error. $\endgroup$ – André Nicolas Nov 9 '13 at 21:15
  • $\begingroup$ @GitGud that's exactly what prompted this question. I got a result that was different than the solution, but I'm not sure if it is wrong. The only difference is that I didn't change my limits of integration. $\endgroup$ – hax0r_n_code Nov 9 '13 at 21:18
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    $\begingroup$ @inquisitor It is wrong in general. See this. $\endgroup$ – Git Gud Nov 9 '13 at 21:19
  • $\begingroup$ @inquisitor : There are some situations in which changing the bounds is not optional. See my answer below. $\endgroup$ – Michael Hardy Nov 9 '13 at 21:33

Sometimes we are doing an indefinite integral by $u$-substitution, and then the form of integral involving $u$ is easily converted back into $x$ or whatever the original variable was at the end of the process.

However when a definite integral is involved, you have a choice of either converting the limits of integration from (say) $x$ limits to $u$ limits, or considering the $u$-substitution as a means to obtaining the final indefinite integral in terms of $x$ and using the original limits of integration.

The former has the advantage of skipping the substitution back into $x$, but at the cost of figuring out how to change the limits of integration into terms of $u$. This was a bad habit you learned, or more precisely, a useful habit you failed to learn. Converting the limits from $x$ to $u$ is ordinarily just a matter of using the $x$ limits in the expression for $u$ in terms of $x$.

  • $\begingroup$ Thank you for your answer. What troubles me is early in our learning integration, the book seemed to indicate a choice as to using the substituted limits of integration, so I just did what I learned first. I'm going to try and break myself out of this habit. $\endgroup$ – hax0r_n_code Nov 9 '13 at 21:24

The idea is that the limits of the integral are in terms of the variable of integration. So take for instance: $$ \int_{0}^{2\pi}\cos(x)e^{\cos x}dx $$ Notice that our function is periodic with period $2\pi$ and symmetric about its half period point, $\pi$; therefore, $$ \int_{0}^{2\pi}\cos(x)e^{\cos x}dx=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx $$ This will be important when we make the $u$-substitution. In this case, we might choose $u=\cos(x)$ with $du=-\sin(x)dx$; therefore, $$ \int_{0}^{2\pi}\cos(x)e^{\cos x}dx\\=2\int_{0}^{\pi}\cos(x)e^{\cos x}dx=-2\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du $$ You'll notice a few things about this process. First, if I had left the limits of integration as $0$ to $\pi$ after making the $u$-substitution, then the two integrals would have been different since the functions being integrated over would have been different. Remember, the integration variable is a dummy variable, so what you use for it is completely irrelevant. Thus, $$ \int_{0}^{2\pi}\cos(x)e^{\cos x}dx=\int_{0}^{2\pi}\cos(u)e^{\cos u}du $$ Second, the way you decide how the limits change is by evaluating the parameterization equation. So for the above example $u(x)=\cos(x)$ so $u(0)=1$ and $u(\pi)=-1$. If I had left the limits as $0$ and $2\pi$, then the $u$-integral would have been from $1$ to $1$ which is obviously wrong since it evaluates to $0$ and the graph of our function clearly has a net positive area over the interval.

Edit: As Barry Cipra alludes to in the comments, there is a neat explanation for why the integral from $0$ to $2\pi$ causes problems when the variable changes. Specifically, I used the following identities in the above calculations $$ \cos(x)=u\\ \sin(x)=\sqrt{1-\cos^2(x)}\\ \therefore \ -\sin(x)=-\sqrt{1-u^2} $$ But this is not always true. The second line comes from $\sin^2(x)+\cos^2(x)=1$, which means that you have to take a square root. And whenever a square root is introduced so too must a plus-or-minus sign. In other words, the correct identity should be $$ -\sin(x)=\mp\sqrt{1-u^2} $$ with the sign chosen based on the value of the input, $x$.

In this problem, $\sin(x)$ is positive from $0$ to $\pi$, but it is negative from $\pi$ to $2\pi$. So the result above could also be derived as follows: $$ \int_{0}^{2\pi}\cos(x)e^{\cos x}dx=-\int_{1}^{-1}\frac{e^{u}u}{\sqrt{1-u^2}}du+\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}}du=2\int_{-1}^{1}\frac{e^{u}u}{\sqrt{1-u^2}} $$

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    $\begingroup$ How it is possible what u(2π)=−1 ? $\endgroup$ – Anton Malmygin May 23 '17 at 16:08
  • $\begingroup$ Thanks for pointing this out. I fixed it. I must have been writing this answer quickly and didn't do the calculation correctly. $\endgroup$ – Geoffrey Jun 26 '17 at 17:25
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    $\begingroup$ I think you should maybe say more about exactly where the error occurs in writing $\int_0^{2\pi}\cos(x)e^{\cos x}dx=\int_1^1{e^uu\over\sqrt{1-u^2}}du=0$. (Hint: Is it always true that $\sin x=\sqrt{1-\cos^2x}$?) $\endgroup$ – Barry Cipra Jun 26 '17 at 17:40
  • $\begingroup$ Okay, that seems like a good idea. Might as well be as complete as possible. $\endgroup$ – Geoffrey Jun 26 '17 at 18:12

Suppose you have $$ \int_0^{\pi/4} \left(\tan x\right)^3 \Big(\sec^2 x\,dx\Big) $$ (where you should take the $\Big($Big parentheses$\Big)$ as a hint about what $u$ should be), then you can write \begin{align} u & = \tan x \\ du & = \sec^2 x\,dx \end{align} and then the integral is $$ \int_{x=0}^{x=\pi/4} u^3 \,du = \left[\frac{u^4}4\right]_{x=0}^{x=\pi/4} = \left[\frac{(\tan x)^4}{4}\right]_0^{\pi/4} = \frac14. $$ In this case, you could say changing the bounds is optional, although it speeds things up and simplifies things: $$ \int_0^1 u^3\,du = \left[ \frac{u^4}{4}\right]_0^1 = \frac14. $$

But there are situations in which you'll get nowhere if you don't change the bounds. For example, the Beta function is defined by $$ B(\alpha,\beta) = \int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx. $$ So suppose we have $$ \int_0^5 x^{\alpha-1}(5-x)^{\beta-1} \,dx. $$ (I've seen integrals like this arising in statistics problems.) Then let \begin{align} u & = x/5, \\ du & = dx/5. \end{align} The integral then becomes $$ \int_0^1 (5u)^{\alpha-1} (5-5u)^{\beta-1}\,\Big(5\,du\Big) = 5^3 \int_0^1 u^{\alpha-1}(1-u)^{\beta-1}\, du = 125 B(\alpha,\beta). $$

Without changing the bounds, how could one recognize this as being the Beta function?


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