Consider the function f  defined by f(x)=x2  on the interval [−2,2] . Let Pn:{−2,−2+ 4 n ,−2+ 8 n ,⋯,−2+ 4n n =2}, i.e. Pn is a partition of [−2,2] by dividing it into n equal subintervals.  Therefore, Δxi= [ Select ] 2/n no enough information 4/n , i=1,2,⋯,n. If we pick x ∗ i ∈[xi−1,xi] such that x ∗ i =xi−1, then x ∗ i = [ Select ] 2(2i-2-n)/n 2(2i+n)/n 2(2i-n)/n 2(2i-2+n)/n . If we pick x ∗ i ∈[xi−1,xi] such that x ∗ i =xi, then x ∗ i = [ Select ] 2(2i+n)/n 2(2i-n)/n 2(2i-2+n)/n 2(2i-2-n)/n . Then we can write the Riemann sum of f and Pn as  S ∗ Pn (f)= n ∑ i=1f(x ∗ i )Δxi.