For an integral $\int^b_af(x)dx$,it i substitute $x$ with $at$, the I have ${1\over a}\int^{b\over a}_1 f(t)dt.$
What confuses me is that why would we have a $1\over a$ front. From the integral's definition point of view, we have already chop $\Delta x$ small enough, why do we have to make the $\Delta t $ $1\over a$times smaller? Is there any intuitive explanat