RAM (Rectangle Approximation Method/Riemann Sum)

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Simple Riemann approximation using rectangles
6:45 — by Khan Academy

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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) curve

    graph{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.)

  • This key question hasn't been answered yet. Answer question
  • 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.

  • 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.

  • #int_1^3 e^x dx approx R_n=sum_{i=1}^n e^(1+iDelta x) Delta x#,

    where #Delta x={3-1}/n=2/n#

    #= sum_{i=1}^n e^{1+2/ni}2/n#

    #={2e}/n sum_{i=1}^{n}(e^{2/n})^i#

    #= {2e}/n{e^{2/n}[1-(e^{2/n})^n]}/{1-e^{2/n}}#

    #={2e^{1+2/n}(1-e^2)}/{n(1-e^{2/n})}#

    you can continue from here.

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