2
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.