I have confused myself with another study question from Apostol "Calculus" Volume 1. It's Section 5.8 Question 21 which states:

Deduce the formulas in Theorem 1.18 and Theorem 1.19 by the method of substitution

For those not familiar with Apostol, I quote the following two relevant Theorems from Apostol:

(1.18) INVARIANCE UNDER TRANSLATION: If $f$ is integrable on $[a, b]$, then for every real $c$ we have $$ \int_a^b f(x) dx = \int_{a+c}^{b+c} f(y - c) dy $$

(1.19): EXPANSION OR CONTRACTION OF THE INTERVAL OF INTEGRATION: If $f$ is integrable on $[a, b]$, then for every real $k\neq 0$ we have $$ \int_a^b f(x) dx = \frac{1}{k} \int_{ka}^{kb} f(\frac{y}{k}) dy $$ For Theorem 1.18 I deduced that the substitution $y = x + 2c$ would be sufficient as it would give the following: $$ \int_a^b f(x) dx = \int_{a+2c}^{b+2c} f(y) dy = \int_{a+c}^{b+c} f(y - c) dy $$ As far as I can tell, this is correct.

For Theorem 1.19 however, I've run into a problem. The substitution $y = kx$ didn't seem to work as it gave me the following: $$ \int_a^b f(x) dx = \frac{1}{k} \int_a^b k f(x) dx = \frac{1}{k} \int_{ka}^{kb} f(y) dy $$ and when I tried $y = k^2 x$ I arrived at the following: $$ \int_a^b f(x) dx = \frac{1}{k^2} \int_a^b k^2 f(x) dx = \frac{1}{k^2} \int_{k^2a}^{k^2b} f(y) dy = \frac{1}{k^2} \int_{ka}^{kb} f(\frac{y}{k}) dy $$ I'm clearly missing the point somewhere on this, so any advice and guidance would be most welcome.

  • 2
    $\begingroup$ $kf(x)\ne f(kx)$! $\endgroup$ – Martín-Blas Pérez Pinilla Mar 19 '14 at 23:39
  • $\begingroup$ I don't understand this: what are you trying to say?\ $\endgroup$ – emjay Mar 19 '14 at 23:43
  • $\begingroup$ That you are doing wrong the substitution $y=kx$. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 19 '14 at 23:46
  • $\begingroup$ Ok, so correct me if Im wrong here: $\endgroup$ – emjay Mar 19 '14 at 23:47
  • $\begingroup$ Sorry, pressed return too early, my bad. I think the question has been answered for me. Consequently, I see where my thinking is off. $\endgroup$ – emjay Mar 19 '14 at 23:56

You are missing the point in substitution of $$ dx = \frac{dy}{k} $$ $$ To \quad prove \\ \int^b_a f(x)dx = \frac{1}{k}\int_{ka}^{kb}f(\frac{y}{k})dy \\ --------------------------------\\ x = \frac{y}{k} \Rightarrow dx = \frac{dy}{k} \quad \because k \in \Re \\ x = a \Rightarrow y = ak \quad || \quad x = b \Rightarrow y = bk \\ \int^b_a f(x)dx = \int_{ka}^{kb}f(\frac{y}{k})(\frac{dy}{k}) \\ \quad = \frac{1}{k}\int_{ka}^{kb}f(\frac{y}{k})dy $$


When $y=kx$ : $$\int_a^b f(x) \operatorname{d}x = \int_{ka}^{kb} f\left(\frac{y}{k}\right) \operatorname{d}\left(\frac{y}{k}\right) = \frac 1 k \int_{ka}^{kb} f\left(\frac{y}{k}\right) \operatorname{d}y $$

When $y=k^2 x$ : $$\int_a^b f(x) \operatorname{d}x = \int_{k^2a}^{k^2b} f\left(\frac{y}{k^2}\right) \operatorname{d}\left(\frac{y}{k^2}\right) = \frac 1 {k^2} \int_{k^2a}^{k^2b} f\left(\frac{y}{k^2}\right) \operatorname{d}y $$


Remember that substituting in an integral means exactly that - you have to substitute (for example) $x$ in terms of $y$, and you have to do it both in the limits of integration and in the integrand, as well as getting $dx$ in terms of $dy$. You appear to be under the impression that you only have to substitute in the limits, so that for any substitution you get something like $$\int_a^b f(x)\,dx=\int_?^? f(y)\,dy\ .$$ For example, to evaluate $$\int_0^2 2x(x^2+5)^7\,dx$$ by substituting $y=x^2+5$ you would have $$\int_0^2 2x(x^2+5)^7\,dx=\int_5^9 2y(y^2+5)^7\,dy\ .$$ This is not correct, the $(x^2+5)^7$ should be just $y^7$, not $(y^2+5)^7$. The full working could be set out like this: $$y=x^2+5\quad\hbox{so}\quad \frac{dy}{dx}=2x$$ and therefore $$\eqalign{\int_0^2 2x(x^2+5)^7\,dx &=\int_0^2 (x^2+5)^7\,\frac{dy}{dx}\,dx\cr &=\int_5^9 y^7\,dy\cr}$$ which is now an easy integral. Hopefully you can now do the integrals from Apostol in a similar way. Hint: substitute $y=x+c$ for the first one.

  • $\begingroup$ Invalid answer. The rules you use here are precisely the sort of thing you are supposed to be proving in the propositions in the question. $\endgroup$ – Frank Mar 19 '14 at 23:52
  • $\begingroup$ @Frank, not at all. The second line of the question says, "Deduce the formulas. . . by the method of substitution". It's not asking for a basic argument from Riemann sums or anything like that. $\endgroup$ – David Mar 19 '14 at 23:54
  • $\begingroup$ Yes, I apologize. I didn't read the question properly. I guess I intuitively thought that these propositions are far more basic than the proof of 'general' integration by substitution, and thus call for a more elementary proof. Sorry once again. Frank $\endgroup$ – Frank Mar 20 '14 at 0:01

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