RAM (Rectangle Approximation Method/Riemann Sum)
Key Questions
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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 right-angled 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 left-hand side of that rectangle. So:
- The rectangle between 0 and 0.7 has height 0, because f(x) = 0 on the left-hand side.
- The rectangle between 0.7 and 1.4 has height 1.05, because f(x) = 1.05 on the left-hand side.
- The rectangle between 1.4 and 2.1 has height 2.1, because f(x) = 2.1 on the left-hand 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 left-hand side. You can choose the right-hand side, or the midpoint, but make sure you do the same thing for the entire question - don't take one rectangle from the left-hand side and one from the right.
Also note: the smaller the rectangles, the more precise your answer!
CJ · · Oct 2 2014
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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={b-a}/n={2-1}/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.
Wataru
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3
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Oct 2 2014
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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={b-a}/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.
Wataru
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Oct 10 2014
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In my opinion, you do not need to unless it takes too long to work out integrals. If integrals are time-consuming, and you do not need an exact value for your purposes, it makes sense to approximate them.
I hope that this was helpful.
Wataru
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1
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Oct 17 2014