Integration by substitution, why do we change the limits? I've highlighted the part I don't understand in red. Why do we change the limits of integration here? What difference does it make?

Source of Quotation: Calculus: Early Transcendentals, 7th Edition, James Stewart

No, we don't change the limits. Who said that. They stay the same. Only their representation changes.
$$
L = \int\limits^{x=4}_{x=1}\sqrt{1 + \dfrac{9}{4}x}dx
$$
Now we substitute $u=1+\dfrac{9}{4}x$:
$$
L = \int\limits^{x=4}_{x=1} \sqrt{u} dx
$$
See? We didn't change anything. We only substituted $u$ for the $x$ expression in the integrand. But now, it contains mixed variables and not solvable as is. We need to substitute $u$ expressions for other $x$ expressions too.
$$
L = \int\limits^{x=4}_{x=1} \sqrt{u} dx = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right)
$$
Now, all the variables are in $u$, and the integral is solvable.
$$
\dfrac{4}{9}(u-1)=4 \implies u=10, \quad \dfrac{4}{9}(u-1)=1 \implies \dfrac{13}{4} \\
d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9}du
$$
We rewrite the integral expression:
$$
L = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9} \int\limits^{u=10}_{u=\dfrac{13}{4}} \sqrt{u} du
$$
Have we ever changed the limits in any step? No! They are the exact same limits. There only represented by $u$ instead of $x$ in the final expression.
