Please consider the integral:


I would like to know why I can't make the substitution:


When I make the substitution, the limits of the integral will be the same, and the integral itself will be zero, which is the wrong answer. So why does this simple change of variables not work as I have expected?

Please note that I do not want help solving the integral, I know how to solve it several ways. My question is why does this specific attempt at a solution not work?

  • $\begingroup$ See Theorem 12.12 of this. Also this may be of interest. $\endgroup$ – user 170039 Nov 12 '16 at 9:21
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    $\begingroup$ Things like this are the stupidest problems you would probably expect to be much more simple, but no, just because $x^2=(-x)^2$, everything breaks down. $\displaystyle\overbrace{\left(\ddot{\stackrel{\quad>}{\frown}}\right)}_{\begin{align}\hline\qquad\end{align}}$ $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 13:43

This is a great question, and I find that variants are often poorly understood. Let's carry out the substitution completely. Since $a$ clearly doesn't matter, I'll suppose that $a = 1$ for simplicity.

$$ \int_{-1}^1 x^2 dx = \frac{1}{2} \int_{-1}^1 x \cdot 2x dx.$$

In this form, we substitute $u = x^2$, so that $du = 2x dx$. Then in the integral, we have

$$ \frac{1}{2} \int_{-1}^1 \underbrace{x}_{\pm\sqrt u} \cdot \underbrace{2x dx}_{du}.$$

To complete the substitution, we need to substitute $x = \pm \sqrt u$. I write $\pm$ to indicate that sometimes we have $x = \sqrt u$ and sometimes we have $x = - \sqrt u$. In particular, when $x$ is negative, we choose the negative square root, $x = - \sqrt u$. This ambiguity is very important.

Namely, for $x$ in $[-1, 0]$, we have that $x = -\sqrt u$ and for $x$ in $[0, 1]$ we have that $x = \sqrt u$. Then to perform the substitution, we split the integral into these two intervals and handle each separately.

$$\begin{align} \frac{1}{2} \int_{-1}^1 x \cdot 2x dx &= \frac{1}{2} \int_{-1}^0 x \cdot 2x dx + \frac{1}{2} \int_{0}^1 x \cdot 2x dx \\ &= \frac{1}{2} \int_1^0 (- \sqrt u) du + \frac{1}{2} \int_0^1 \sqrt u du \\ &= \frac{1}{2} \int_0^1 \sqrt u du + \frac{1}{2} \int_0^1 \sqrt u du \\ &= \int_0^1 \sqrt u du = \frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{2}{3}. \end{align}$$

The exact nature of the substitution is very important. This causes some introductory texts to state that substitutions must be injective. But a careful analysis shows otherwise. $\spadesuit$

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    $\begingroup$ Thanks very much. I think this is one of those examples of me learning a topic in high school, which was not a fully rigorous treatment, and then going through the next 10 years thinking I understood it in full detail when really I did not! Thanks again. $\endgroup$ – Lachy Nov 12 '16 at 8:32
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    $\begingroup$ I don't even understand why you (and others) present substitution this way. This isn't even a substitution to me; it's a clever symbolic manipulation. A substitution would involve substituting something for a variable, in this case $x$. So if you want $u = x^2$, then you have to solve for $x$ and then substitute that into the integral, which incidentally forces you to determine the sign of the square root at the first step: $x = \sqrt{u}$, $dx = \frac{du}{2\sqrt{u}}$ etc... $\endgroup$ – Mehrdad Nov 12 '16 at 11:33
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    $\begingroup$ @mixedmath: your substitution is injective; that's why you can do what you did. Substitution comes from the equality $$\int_a^bf (x)dz=\int_{g (a)}^{g (b)} f (g (t))\,g'(t)\,dt, $$which requires $g $ to be injective on $[a,b] $ to hold. $\endgroup$ – Martin Argerami Nov 12 '16 at 12:13
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    $\begingroup$ @Mehrdad Sometimes it is about which way comes out better for the audience. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 13:39
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    $\begingroup$ @mixedmath: but what you are doing is splitting your integral in two halfs, and in each if them your substitution is injective. $\endgroup$ – Martin Argerami Apr 7 '17 at 16:38

I like Alfred's point but I would say it slightly differently. The $dx$ at the end is not just the “other end” of the integral delimited on the left by $\int$. It's also not just the indicator of which variable is being integrated. It's part of the integral itself. In other words, we don't integrate functions $f(x)$ so much as forms $f(x)\,dx$.

When we make a substitution, we have to substitute fully the entire form from one involving $x$ to one involving $u$. When integrating $x^2\,dx$, if we substitute $u = x^2$, we must substitute for $dx$ in terms of $u$ and $du$. Since $\frac{du}{dx} = 2x$, we write $du = 2x\,dx$. Thus $dx = \frac{1}{2x}\,du = \frac{1}{2\sqrt{u}}\,du$. So $$ \int x^2 \,dx = \int u \frac{1}{2\sqrt{u}}\,du = \frac{1}{2}\int \sqrt{u}\,du $$ That's different from $\int u\,du$.

I think you were just switching the differential from $dx$ to $du$ without substituting. You can avoid this mistake if you remember that $dx$ and $du$ are part of the integral.

  • $\begingroup$ -1: for two reasons. The "integral of a form" point of view might fit some areas of math, but it is definitely orthogonal to many others ( I don't see how you would describe the dual of $C [0,1] $ by thinking of integrals as "integrals of forms"). More importantly, this does not answer the question; the OP clearly knows how to do substitution, and is asking what happens with the limits in his example. $\endgroup$ – Martin Argerami Nov 12 '16 at 12:02

It´s supposed, when using substitution to evaluate a definite integral, that you reverse substitution before apply limits. As you choose to substitute $u = x^2$ and so $du = 2x dx$, you have to "adjust" the integral before perform that substitution:

$$\int_{-a}^a x^2 dx = \frac{1}{2} \int_{-a}^a \underbrace{x}_{u^{1/2}} \cdot \underbrace{2x dx}_{du} $$

As it´s planned to reverse the substitution, it´s better avoid to adjust the limits of the integral and evaluate it as an indefinite integral:

$$\begin{align} \frac{1}{2}\int u^{1/2} du=\frac{1}{2}.\frac{2}{3} u^{3/2} + C=\frac{1}{3} u^{3/2} + C \end{align}$$

Reversing substitution, i.e. $u = x^2$, and applying limits:

$$\begin{align} =\frac{1}{3} {(x^2)}^{3/2}\bigg|_{-a}^a =\frac{1}{3} x^3 \bigg|_{-a}^a =\frac{1}{3} a^3 -\frac{1}{3} {(-a)}^3 =\frac{2}{3} a^3 \end{align}$$

  • $\begingroup$ It's interesting that in the end this comes out to the same value as if we did the integral without any attempt at substitution, because some of its intermediate steps are not right. For one thing, the interval of integration is not adjusted for the substitution; if $u=x^2$ then setting $u=a$ is not the same thing as setting $x=a$; and setting $u=-a$ does not work at all within real analysis. $\endgroup$ – David K Nov 12 '16 at 23:17
  • $\begingroup$ @DavidK , thank you for your comment. The reason why it comes out the same value is because I didn't replaced the limits on u.They were replaced after sbstitution was reversed to x. But you are right on mentioning those limits were not adjusted when substitution was done, so I edit my post to better affording, using, now, an undefined integral. $\endgroup$ – elcio.humphreys Nov 15 '16 at 20:02
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    $\begingroup$ It's still hazardous to apply indefinite integrals outside of their domains. And $(x^2)^{3/2} = x^3$ only when $x$ is positive. Altogether I think it's really far better to split the integral into two parts, one for $[-a,0]$ and one for $[0,a]$, at the very beginning and avoid all these pitfalls. $\endgroup$ – David K Nov 15 '16 at 20:10

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