We'll use right Riemann sums with a=0 and b=2. When there are n rectangles,
Δx=nb−a=n2 and xi=a+iΔx=2i/n. So
we need to evaluate
n→∞limi=1∑nf(xi)Δx=n→∞limi=1∑n((xi)3+xi)Δx=n→∞limi=1∑n((n2i)3+n2i)n2=n→∞limn2i=1∑n(n38i3+n2i)=n→∞lim(n416i=1∑ni3+n24i=1∑ni)=n→∞lim(n4⋅416(n4+2n3+n2)+n2⋅24(n2+n))=n→∞lim(416(1+n2+n21)+24(1+n1))=416+24=6.