We divide the interval

$[1,3]$

into $4$ subintervals, each of width

$12$

$.$

The first subinterval has left endpoint $1.$

Find the remaining left endpoints of the subintervals.

The second subinterval has left endpoint

$1+1$

$12$

$=$

$2$

$.$

The third subinterval has left endpoint

$1+2$

$12$

$=$

$2$

$.$

The fourth $($last$)$ subinterval has left endpoint

$1+3$

$12$

$=$

$2$

$.$