Use Riemann sums to evaluate? : #int_0^3 \ x^2-3x+2 \ dx #

We are asked to evaluate:

# I = int_0^3 \ x^2-3x+2 \ dx #

Using Riemann sums. By definition of an integral, then

# int_a^b \ f(x) \ dx #

represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#

Here we have #f(x)=x^2-3x+2# and we partition the interval #[0,3]# using:

# Delta = {0, 1*3/n, 2*3/n, ..., n*3/n } #

And so:

# I = int_0^3 \ (x^2-3x+2) \ dx #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(0+i*(3-0)/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f((3i)/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {((3i)/n)^2-3((3i)/n)+2}#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(9i^2)/n^2-(9i)/n+2}#

# \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n (9i^2)/n^2 - sum_(i=1)^n (9i)/n + sum_(i=1)^n 2}#

# \ \ = lim_(n rarr oo) 3/n {9/n^2 sum_(i=1)^n i^2 - 9/n sum_(i=1)^n i + 2 sum_(i=1)^n 1}#

Using the standard summation formula:

# sum_(r=1)^n r \ = 1/2n(n+1) #
# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

we have:

# I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1) - 9/n 1/2n(n+1) + 2 n}#

# \ \ = lim_(n rarr oo) 3/n {3/(2n) (n+1)(2n+1) - 9/2 (n+1) + 2 n}#

# \ \ = lim_(n rarr oo) 3/n 1/(2n) {3 (n+1)(2n+1) - 9n (n+1) + 4n^2}#

# \ \ = lim_(n rarr oo) 3/(2n^2) {3 (2n^2+3n+1) - 9(n^2+1) + 4n^2}#

# \ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3 -9n^2-9 + 4n^2}#

# \ \ = 3/2 lim_(n rarr oo) 1/(n^2) {n^2+9n -6}#

# \ \ = 3/2 lim_(n rarr oo) {1+9/n -6/n^2}#

# \ \ = 3/2 {1+0 -0}#

# \ \ = 3/2#

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# I = int_0^3 \ x^2-3x+2 \ dx #
# \ \ = [x^3/3-(3x^2)/2+2x]_0^3 #
# \ \ = (9-27/2+6) - (0) #
# \ \ = 3/2 #