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CalcDerp102

  • 2 years ago

Im doing a problem at the moment and im stuck its a right riemann sum problem, equation and how far i got posted below

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  1. CalcDerp102
    • 2 years ago
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    \[\int\limits_{3}^{0} (2x^2 + x +3) dx\] \[\Delta x = -3/n\] \[x_{k} = 3 + -3k/n\] \[f(x_k) = (3+ -3k/n)^2 - (3 + -3k/n) +1\]

  2. CalcDerp102
    • 2 years ago
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    im not sure where to go from there

  3. Algebraic!
    • 2 years ago
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    *

  4. Algebraic!
    • 2 years ago
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    I think you'd be better off using: \[\huge \int\limits_{a}^{b} f(x) dx = -\int\limits_{b}^{a} f(x) dx\] at the start

  5. Algebraic!
    • 2 years ago
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    then \[f(c _{k}) = 2*k^2 *\Delta x ^3 + k*\Delta x^2 + 3 \Delta x\]

  6. Algebraic!
    • 2 years ago
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    and \[S _{n} = 2 \Delta x ^3 \sum_{}^{} k^2 + \Delta x^2 \sum_{}^{} k +3 \Delta x\]

  7. CalcDerp102
    • 2 years ago
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    so if be better of making the original equation negative then using that formula to find the equation

  8. Algebraic!
    • 2 years ago
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    a bit easier, yes... that's probably why they gave it to you in the form that they did... so you could make the easy simplification

  9. Algebraic!
    • 2 years ago
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    you got it from here?

  10. CalcDerp102
    • 2 years ago
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    yea i think so i think what confuses me the most is 3+(-3k/n)

  11. Algebraic!
    • 2 years ago
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    do what I did before you sub.s in 'delta x = b/n'

  12. CalcDerp102
    • 2 years ago
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    \[2 * (3+ \frac{ -3k }{ n }^2) * \frac{ -3 }{ n }^3 + (3+ \frac{ -3k }{ n }) * \frac{ -3 }{ n } + 3\frac{ -3 }{ n }\] is this what im looking for or do does k = 1

  13. Algebraic!
    • 2 years ago
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    start at x=0... move delta x to the right... what's the height of the rectangle you'd draw there?

  14. Algebraic!
    • 2 years ago
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    |dw:1354238714258:dw|

  15. Algebraic!
    • 2 years ago
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    height is f(delta x) width is delta x area is f(delta x) * delta x which is (2*(delta x)^2 + delta x +3))*delta x

  16. Algebraic!
    • 2 years ago
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    |dw:1354238854668:dw|

  17. Algebraic!
    • 2 years ago
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    next step: go another 'delta x' to the right height is 2*(2*delta x)^2 + 2*delta x +3 width is delta x area is (2*(2*delta x)^2 + 2*delta x +3)*delta x

  18. Algebraic!
    • 2 years ago
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    so the sum of areas is going to look like: (2*(k*delta x)^2 + k*delta x +3)*delta x

  19. Algebraic!
    • 2 years ago
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    2*k^2*(delta x)^3 + k*(delta x)^2 +3*delta x

  20. Algebraic!
    • 2 years ago
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    summing from k=1 to n

  21. Algebraic!
    • 2 years ago
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    that's the sum I wrote above

  22. CalcDerp102
    • 2 years ago
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    oh alright, thanks i appreciate the help!

  23. Algebraic!
    • 2 years ago
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    does that make sense?

  24. CalcDerp102
    • 2 years ago
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    Yeah it does, im trying to work through the problem. ill have an answer soon enough though

  25. Algebraic!
    • 2 years ago
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    still there?

  26. Algebraic!
    • 2 years ago
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    I didn't write the last term on the summation properly

  27. Algebraic!
    • 2 years ago
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    it should be:\[\Delta x \sum_{k=1}^{n} 3\]

  28. Algebraic!
    • 2 years ago
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    so\[S _{n} = 2(\Delta x)^3\sum_{k=1}^{n} k^2 + (\Delta x)^2 \sum_{k=1}^{n} k +\Delta x \sum_{k=1}^{n} 3\]

  29. CalcDerp102
    • 2 years ago
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    oh ok, thats makes more sense now

  30. CalcDerp102
    • 2 years ago
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    Its either 0 or -6 i think 0 because n = 3, am i correct?

  31. Algebraic!
    • 2 years ago
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    it's -63/2

  32. Algebraic!
    • 2 years ago
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    n is the number of divisions in the partition... you find the sum above (in terms of n) ... you sub.s in b/n for delta x and then you take the limit as n->infinity

  33. Algebraic!
    • 2 years ago
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    then use b=3 (the interval length) and finally, make the whole result negative ( due to the simplification we did at the very beginning)

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