RAM (Rectangle Approximation Method/Riemann Sum)
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Key Questions

Not every function (and not every interesting and important function) has an antiderivative that is finitely expressible using the algebraic operations: addition, subtraction, multiplication, division and extraction of roots.
Two examples:
Natural Logarithm
#int_1^b 1/x dx# The natural log must be approximated using some approximation technique  by approximating the integral or by some series approximation.
Probablity and Statistics
The standard Normal (or Bell or Gaussian) curvegraph{e^(1/2 x^2)/sqrt(2pi) [2.398, 2.469, 0.55, 1.883]}
#1/sqrt(2pi) int_0^z e^(1/2 x^2)dx# gives the probability of a random variable having a standard normal value between#0# and#z# This integral cannot be expressed finitely using algebraic operations and must be approximated numerically. (As must
#e^x# itself.) 
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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={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.

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