The area A of the region under the graph of #f# above the #x#-axis from #x=a# to #b# can be found by

#A=lim_{n to infty}sum_{i=1}^n f(x_i) Delta x#,

where #x_i=a+iDelta x# and #Delta x={b-a}/n#.

Let us find the area of the region under the graph of #y=2x+1# from #x=1# to #3#.

By definition,

#A=lim_{n to infty}sum_{i=1}^n[2(1+2/ni)+1]2/n#

by simplifying the expression inside the summation,

#=lim_{n to infty}sum_{i=1}^n(8/n^2i+6/n)#

by splitting the summation and pulling out constants,

#=lim_{n to infty}(8/n^2sum_{i=1}^ni+6/nsum_{i=1}^n1)#

by the summation formulas #sum_{i=1}^ni={n(n+1)}/2# and #sum_{i=1}^n1=n#,

#=lim_{n to infty}(8/n^2cdot{n(n+1)}/2+6/ncdot n)#

by cancelling out #n#'s,

#=lim_{n to infty}[4(1+1/n)+6]=4(1+0)+6=10#

I hope that this was helpful.