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. 

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