I'm going to assume that you want to evaluate the integral, not simply get an approximation.
You didn't specify, so I'll use right endpoint in my calculations.
Also, you may be required to write equalities all the way through, for students learning this, I think it's more clear to evaluate the pieces, then the sum and finally evaluate the limit.
#int_1^3 (2x+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x#
For each #n#, we have:
#Delta x = (b-a)/n = 2/n#
#x_i = a+iDeltax=1+(2i)/n #
#f(x_i) = 2x_i+1 = 2(1+(2i)/n)+1=(4i)/n+3#
#sum_(i=1)^n f(x_i) Delta x = sum_(i=1)^n ((4i)/n+3) 2/n = sum_(i=1)^n ((8i)/n^2+6/n)#
So the sum is:
#sum_(i=1)^n f(x_i) Delta x = sum_(i=1)^n (8i)/n^2+sum_(i=1)^n 6/n#.
And finally:
#sum_(i=1)^n f(x_i) Delta x = 8/n^2sum_(i=1)^n i + 6/n sum_(i=1)^n 1#
Apply the formulas for these sums to get:
#sum_(i=1)^n f(x_i) Delta x = 8/n^2 (n(n+1))/2 + 6/n (n)#.
In preparation for evaluating the limit as #nrarroo#, rewrite this as:
#sum_(i=1)^n f(x_i) Delta x = (4 n)/n (n+1)/n + (6n)/n = 4(1+1/n)+6#
Evaluate the limit as #nrarroo#:
#int_1^3 (2x+1)dx = lim_(nrarroo) sum_(i=1)^n f(x_i) Delta x#
#color(white)"ssssssssssssssss"# # = lim_(nrarroo) sum_(i=1)^n ((4i)/n+3) 2/n#
#color(white)"ssssssssssssssss"# # = lim_(nrarroo) (4(1+1/n)+6)#
#color(white)"ssssssssssssssss"# # = 4+6 = 10#
#int_1^3 (2x+1)dx = 10#
