## University Calculus: Early Transcendentals (3rd Edition)

$1.2$
We are given: $0,1.2,1.5,2.3,2.6,3$ We need to split the given points in the intervals as follows: For sub-interval $[0,1.2]$, we have $1.2-0=1.2$; For sub-interval $[1.2,1.5]$, we have $1.5-1.2=0.3$ For sub-interval $[1.5,2.3]$, we have $2.3-1.5=0.8$ For sub-interval $[2.3,2.6]$, we have $2.6-2.3=0.3$ For sub-interval $[2.6,3]$, we have $3-2.6=0.4$ Thus, the norm of the partition will be the widest interval: $1.2$