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Area between two graphs as a sum

  1. Apr 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap1.jpg
    2. Relevant equations
    $$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$

    3. The attempt at a solution
    $$S_n=2\{[f_1(x_1)-f_2(x_1)]\Delta x+[f_1(x_2)-f_2(x_2)]\Delta x+...+[f_1(x_{n-1})-f_2(x_{n-1})]\Delta x\}$$
    $$S_n=2\{[18-\Delta x^2-\Delta x^2]+[18-(2\Delta x)^2-(2\Delta x)^2]+...+[18-(n-1)^2\Delta x^2-(n-1)^2\Delta x^2]\}$$
    $$S_n=2\{\Delta x[(n-1)18-2(1^2+2^2+...+(n-1)^2)\Delta x]\}$$
    If i develop exactly according to the formula in the Relevant Equations then it's wrong
     
  2. jcsd
  3. Apr 29, 2017 #2

    vela

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    Three questions:
    1. Why the factor of 2 in front?
    2. Why only n-1 terms? Aren't there n rectangles?
    3. Did you take into account the fact that ##\Delta x## depends on ##n##?
    Also, I think you dropped a factor of ##\Delta x## on the second term.
     
  4. Apr 30, 2017 #3

    scottdave

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    It would be nice to know what Eq7 is (perhaps it is your summation equation?). What is the area of interest? The two curves intersect at x = -3 and x = +3.
    Yes as @vela mentioned, delta x would be equal to the total x range, divided by n.
     
  5. Apr 30, 2017 #4
    I calculate only the right half and double the result.
    I calculate only the right half and double the result.
    Snap1.jpg The rectangle at the last ##~\Delta x##, between x3 and x4 is a line, it has area 0, but if you wish....
    $$\frac{1}{2}S_n=[18-\Delta x^2-\Delta x^2]+[18-(2\Delta x)^2-(2\Delta x)^2]+...+[18-(n\Delta x)^2-(n\Delta x)^2]$$
    $$\Delta x=\frac{3}{n}$$
    $$\frac{1}{2}S_n=\Delta x[18n-(1^2+2^2+...+n^2)]=...=\frac{9(n+\frac{1}{5})(n-\frac{1}{2})}{2n^2}~\rightarrow~\frac{9}{2}$$
    While the area, on both sides of the y axis, totals to 104
     
  6. Apr 30, 2017 #5

    vela

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    Your set-up (in the first post) is fine, but your expression for ##S_n## is wrong. All I can say is check your algebra.

    You might want to recheck this value as well. I get 72.
     
  7. May 1, 2017 #6
    $$\frac{1}{2}S_n=\Delta x[18n-2\Delta x^2(1^2+2^2+...+n^2)]=\frac{9}{n^2}(4n^2-3n-1)$$
    Embarrassing but i don't know to solve it. the roots are 2 and -1/2 but when i substitute them ##~4n^2-3n-1\neq 0##
    $$\int_{-3}^3 18-x^2-\int_{-3}^3 x^2=\int_{-3}^3 18=18x\vert _{-3}^3=18\cdot 6=104$$
     
    Last edited: May 1, 2017
  8. May 1, 2017 #7
    Correction:
    $$\frac{1}{2}S_n=\Delta x[18n-2\Delta x^2(1^2+2^2+...+n^2)]=\frac{9}{n^2}(4n^2-3n-1)=\frac{9}{n^2}4(x+\frac{1}{4})(x-1)$$
    $$\frac{1}{2}S_n~\rightarrow~36, ~~S_n~\rightarrow~72$$
    $$\int_{-3}^3 18-x^2-\int_{-3}^3 x^2=\int_{-3}^3 18-2x^2=72$$
    Thank you Vela
     
    Last edited: May 1, 2017
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