RAM (Rectangle Approximation Method/Riemann Sum)
Key Questions

Usually, integration using rectangles is the first step for learning integration. At its most basic, integration is finding the area between the x axis and the line of a function on a graph  if this area is not "nice" and doesn't look like a basic shape (triangle, rectangle, etc.) that we can easily calculate the area of, a good way to approximate it is by using rectangles.
Let's take an example:
#int_0^7(3x)/2dx# This is a function that finds the area between x=0 and x=7 underneath the line of
#f(x)=(3x)/2# . We already know that this is simply going to be a rightangled triangle with base 7 and height 10.5, so the area is going to be#(7*10.5)/2 = 73.5/2=36.75# .Imagine, now, that we didn't have a formula for calculating the area of a triangle, but we did have a formula for calculating the area of a rectangle (which is base x height, as usual). Imagine that we want to "guess" the area of that triangle. So we draw ten rectangles, each with base 0.7. The height is whatever the value of x is at the lefthand side of that rectangle. So:
 The rectangle between 0 and 0.7 has height 0, because f(x) = 0 on the lefthand side.
 The rectangle between 0.7 and 1.4 has height 1.05, because f(x) = 1.05 on the lefthand side.
 The rectangle between 1.4 and 2.1 has height 2.1, because f(x) = 2.1 on the lefthand side.
 And so on.
Eventually, we'll get ten rectangles. We can easily calculate the areas of these rectangles:
 The first rectangle has area
#0.7 * 0 = 0# . Remember that the base of all these rectangles is 0.7  The second rectangle has area
#0.7 * 1.05 = 0.735# .  The third rectangle has area
#0.7 * 2.1 = 1.47# .  And so on.
The areas of the ten rectangles, in order, are:
#0, 0.735, 1.47, 2.205, 2.94, 3.675, 4.41, 5.145, 5.88, 6.615# To get the total area of all the rectangles, we just add them together to get
#33.705# . This is a pretty good approximation for our area above.Note: you don't always have to choose the lefthand side. You can choose the righthand side, or the midpoint, but make sure you do the same thing for the entire question  don't take one rectangle from the lefthand side and one from the right.
Also note: the smaller the rectangles, the more precise your answer!

The interval
#[1,2]# is divided into 5 equal subintervals#[1,1.2],[1.2,1.4],[1.4,1.6],[1.6,1.8], and [1.8,2]# .Each interval are of length
#Delta x={ba}/n={21}/5=0.2# .The midpoints of the above subintervals are
#1.1,1.3,1.5,1.7, and 1.9# .Using the above midpoints to determine the heights of the approximating rectangles, we have
#M_5=[f(1.1)+f(1.3)+f(1.5)+f(1.7)+f(1.9)]Delta x# #=(1/1.1+1/1.3+1/1.5+1/1.7+1/1.9)cdot 0.2 approx 0.692# By Midpoint Rule,
#int_1^2 1/x dx approx 0.692# .I hope that this was helpful.

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={ba}/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.

In my opinion, you do not need to unless it takes too long to work out integrals. If integrals are timeconsuming, and you do not need an exact value for your purposes, it makes sense to approximate them.
I hope that this was helpful.