Because no #n# (number of rectangles) nor choice of sample points is specified, I will assume that we want the area, not just an approximation.
We'll need: #a=1#, #b=2#. We'll use right endpoints as our sample points. (Because the textbook I teach out of does it that way, so I'm used to it.)
For each #n#, we have #Deltax=(b-a)/n = (2-1)/n = 1/n#
The right endpoints of the subintervals are:
#1+1/n, 1+2/n, 1+3/n, . . . 1+i/n, . . . 1+n/n=2#
We are using #x_i"*"=x_i# and #f(x)=3x^2-2#, so
#f(x_i"*") =f(x_i)= 3(x_i)^2-2 = 3(1+i/n)^2-2=3+6i/n+3i^2/n^2-2#,
that is:
#f(x_i"*") = 1+6i/n+3i^2/n^2#. Remember, we have #Deltax=1/n#
Now, we'll use the definition of area or definite integral:
#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i"*") Deltax#
#int_1^2 (3x^2-2)dx = lim_(nrarroo) sum_(i=1)^n (1+6i/n+3i^2/n^2)(1/n)#
#color(white)"sssssssssss"# #= lim_(nrarroo) sum_(i=1)^n (1/n+6i/n^2+3i^2/n^3)#
#color(white)"sssssssssss"# #= lim_(nrarroo) sum_(i=1)^n 1/n+sum_(i=1)^n 6i/n^2+sum_(i=1)^n 3i^2/n^3#
#color(white)"sssssssssss"# #= lim_(nrarroo) 1/nsum_(i=1)^n 1+6/n^2sum_(i=1)^n i+3/n^3sum_(i=1)^n i^2#
Using the three appropriate summation formulas gets us:
#=lim_(nrarroo)(1/n[n]+6/n^2[(n(n+1))/2] +3/n^3 [(n(n+1)(2n+1))/6])#
#= lim_(nrarroo) (1+6/2[(n(n+1))/n^2] +3/6 [(n(n+1)(2n+1))/n^3])#
#color(white)"sssssss"# #= 1+3[1]+1/2[2] = 5#
So,
The area, #int_1^2 (3x^2-2)dx# is #5# (square units)
