Prove that $\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt = \int_{1}^{ab} \frac 1t dt$ Prove that
$$\int_{1}^{a} \frac 1t dt + \int_{1}^{b} \frac 1t dt = \int_{1}^{ab} \frac 1t dt$$
Useful facts:  $\int_{1}^{a}  \frac 1t dt$ can be written as $\int_{b}^{ab} \frac 1t dt$


Every partition $P=${$t_0,...,t_n$} on $[1,a]$ gives rise to a partition $P'=${$bt_0,...,bt_n$} on $[b,ab]$, and conversely.
I have to do the question without actually integrating it, since the Fundamental Theorem of Calculus was not proven yet.

Since $\int_1^a \frac{1}{t}\, dt = \int_b^{ab} \frac{1}{u}\, du$ (using the $u$-sub $u = tb$), we have $$\int_1^a \frac{1}{t}\, dt + \int_1^b \frac{dt}{t} = \int_b^{ab} \frac{1}{u}\, du + \int_1^b \frac{1}{t}\, dt = \int_1^b \frac{1}{t}\, dt + \int_b^{ab} \frac{1}{t}\, dt = \int_1^{ab} \frac{1}{t}\, dt.$$
