Setup the limit definition of definite integrals #int_-1^2(1+x^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. The limit of which is known as the Riemann sum (or the limit definition of an integral)

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)#

And we partition the interval #[a,b]# equally spaced using:

# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } #

# \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b-1 } #

Here we have #f(x)=1+x^2# and so we partition the interval #[-1,2]# using:

# Delta = {-1, -1+1(3)/n, -1+2 (3)/n, -1+3 (3)/n, ..., 2 } #

And so:

# I = int_(-1)^2 \ (1+x^2) \ dx #

# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(-1+(3i)/n) #

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

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

Which is the desired limit.

We can continue and evaluate the limit (ie the integral) by 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) #

So, we have:

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

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

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

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

# \ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(n+1)(2n+1) - 6n(n+1) } #

# \ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(2n^2+3n+1) - (6n^2+6n) } #

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

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

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

# \ \ = 3/2 \ {4 + 0+0 } #

# \ \ = 6 #

Using Calculus

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

# I = int_(-1)^2 \ 1+x^2 \ dx #
# \ \ = [x+x^3/3]_(-1)^2 #
# \ \ = (2+8/3)-(-1-1/3) #
# \ \ = 6 #