Imagine we have:

$$ I_1 = \int_0^x t\,dt $$

if we evaluate this we get:

$$ I_1 = \left[\frac{1}{2} t^2\right]^x_0 $$

$$ = \frac{1}{2}x^2 $$

Lets try substituting in $r = 2t$

Then:

$$ dr/dt = 2 $$

And:

$$ dt = \frac{1}{2}dr $$

So:

$$ I_1 = \int_0^x \left( \frac{1}{2} r \right) \left( \frac{1}{2}\,dr \right) $$

$$ = \int_0^x \frac{1}{4} r\,dr $$

Clearly, this wont give the same answer. So, we probably should modify the limits too.

Try modifying the limit:

$$ x \mapsto 2x $$

So:

$$ I_1 = \int_0^{2x} \frac{1}{4} r \, dr $$

$$ = \left[ \frac{1}{8} r^2 \right]_0^{2x} $$

$$ = \frac{1}{8}(4)x^2 $$

$$ = \frac{1}{2} x^2 $$

Which matches the earlier result.

More generally, looking at https://en.wikipedia.org/wiki/Integration_by_substitution , we hvae:

"Let $I \subseteq \mathbb{R}$ be an interval and $\phi : [a, b] \rightarrow I$ be a differential function with integrable derivative. Suppose that $f:I \rightarrow \mathbb{R}$ is a continuous function. Then

$$ \int_{\phi(a)}^{\phi(b)} f(x)\,dx = \int_a^b f(\phi(t))\phi'(t) \, dt $$

"In other notation: the substitution $x = \phi(t)$ yields:

$$ \frac{dx}{dt} = \phi'(t) $$

"and thus, formally $dx = \phi'(t)\,dt$, which is the required substitution for $dx$

For the substitution of the limits, it looks like we can make it easier to figure out by adding in some additional notation:

$$ \int_{\phi(a)}^{\phi(b)} f(x) \, dx = \int_{x=\phi(a)}^{x=\phi(b)} f(x)\,dx $$

So, substituting $x = \phi(t)$, for the first limit we have:

$$ x = \phi(a) $$

And we want to convert the limits to be in terms of $t$, ie $t_1=$ (something). If $x_1 = \phi(t)$, and we have $x= \phi(a)$

... then $t_1 = a$. Similarly, if $x_2 = \phi(b)$, then $t_2 = b$. As required.

Let's try this notation for the earlier integral $I_1$:

$$ I_1 = \int_0^x t\, dt $$

Substitute $r = 2t$, so we have:

$$ I_1 = \int_0^x t \, dt = \int_{t=0}^{t=x} t \, dt $$

Looking at the limits, which we can denote as $t_1$ and $t_2$, we have:

$$ t_1 = 0, \text{ and }r = 2t $$

Therefore:

$$ r_1 = 2 t_1 = 0 $$

For the upper limit, $t_2$, we have:

$$ t_2 = x, \text{ and }r = 2t $$

Therefore:

$$ r_2 = 2 t_2 = 2x $$

So, $I_1$ will become:

$$ I_1 = \int_{0}^{2x} \left(\frac{1}{2} r\right) \left(\frac{1}{2} \, dr \right) $$

$$ = \left[ \frac{1}{4} \frac{1}{2} r^2 \right]_0^{2x} $$

$$ = \left[ \frac{1}{8} r^2 \right]_0^{2x} $$

$$ = \frac{1}{2} x^2 $$

... as required