Question on Integration by substitution 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. 
 A: You are missing the point in substitution of 
$$
dx = \frac{dy}{k}
$$
$$
To \quad prove \\
\int^b_a f(x)dx = \frac{1}{k}\int_{ka}^{kb}f(\frac{y}{k})dy \\
--------------------------------\\
x = \frac{y}{k} \Rightarrow dx = \frac{dy}{k} \quad \because k \in \Re    \\
x = a \Rightarrow y = ak \quad || \quad x = b \Rightarrow y = bk \\
\int^b_a f(x)dx = \int_{ka}^{kb}f(\frac{y}{k})(\frac{dy}{k}) \\
\quad = \frac{1}{k}\int_{ka}^{kb}f(\frac{y}{k})dy
$$
A: 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 $y=k^2 x$ :
$$\int_a^b f(x) \operatorname{d}x = \int_{k^2a}^{k^2b} f\left(\frac{y}{k^2}\right) \operatorname{d}\left(\frac{y}{k^2}\right) = \frac 1 {k^2} \int_{k^2a}^{k^2b} f\left(\frac{y}{k^2}\right) \operatorname{d}y $$
A: 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.
