Help to understand this property: $\int\limits_{ka}^{kb}s\left(\frac{x}{k}\right)dx = k\int\limits_a^bs(x)dx$ I'm reading the part explaining the properties of the integral of a step function in Apostol's Calculus I and he explains this property:
$$\int\limits_{ka}^{kb}s\left(\frac{x}{k}\right)dx = k\int\limits_a^bs(x)dx$$
by saying that if we distort the horizontal direction (say, the length) by a $k > 0$, it is the same as multiplying the integral by $k$. Intuitively it makes sense: if the area of a rectangle is $\text{Length} \cdot \text{Height}$, then 
$$(k\cdot \text{Length}) \cdot \text{Height} = k\cdot(\text{Length} \cdot \text{Height})$$
But I have some troubles in understanding the form this property takes with ''trickier'' stretching of the interval of integration. I have been playing with the symbol $\int\limits_a^bs(x)dx$ since then, but I'm not sure whether what I did is right. For instance:
Would:
$$
\begin{align*}
&1. \qquad \int\limits_{ka}^{kb}s(x)dx = k\int\limits_a^bs\left(\frac{x}{k}\right)dx\qquad \text{?}\\
&2. \qquad \int\limits_{\sqrt{a}}^{\sqrt{b}}s(x)dx = \left[\int\limits_a^bs(x^2)dx\right]^{1/2}\qquad \text{?}\\
&3. \qquad \int\limits_{a^2}^{b^2}s(x)dx = \left[\int\limits_a^bs(\sqrt{x})dx\right]^{2}\qquad \text{?}\\
&4. \qquad \int\limits_{a/k}^{b/k}s(x)dx = \frac{1}{k}\int\limits_a^bs(kx)dx\qquad \text{?}
\end{align*}
$$
In each case what I did was the following:
Let take $2.$ as an example. If $\sqrt{a} < x < \sqrt{b} \implies a < x^2 < b \implies x^2$ is in the domain of $s$. Then the integrand is $s(x^2)$ on $[a,b]$ and the stretching of the interval (the square root) ''drops'' to the whole integral: $\left[\int\limits_a^bs(x^2)dx\right]^{1/2}$.
If this is correct, then mechanically I know how it works but I'm not able to explain why (in particular, that part where the stretching of $[a,b]$ drops to the integral).
Thanks!!

Hint: Make the substitution $u = \frac{x}{k}$ then $du = \frac{1}{k} dx \implies kdu = dx$.
