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bmelyk

  • 3 years ago

Express the limit as a definite integral over the indicated interval [a,b]

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  1. bmelyk
    • 3 years ago
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    \[\lim n->0 \sum_{i=1}^{n} [2(1+\frac{ 2i }{ n }-5]\frac{ 2 }{ n }\] over [-1,2]

  2. bmelyk
    • 3 years ago
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    lim as n approaches 0 * if that doesnt look clear. looking for a step by step explanation.

  3. Jonask
    • 3 years ago
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    is it like this \[\huge \lim_{n \rightarrow 0} \sum_{i=1}^{n}[2(1+\frac{2i}{n})-5]\frac{2}{n}\]

  4. bmelyk
    • 3 years ago
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    yes.

  5. bmelyk
    • 3 years ago
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    @Jonask

  6. Jonask
    • 3 years ago
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    we want to express the sum as\[\huge \sum_{i=1}^{n}f(c_i)\Delta x\] where \[\Delta x=\frac{2}{n}\]

  7. bmelyk
    • 3 years ago
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    okay.

  8. Jonask
    • 3 years ago
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    we also wanna find c_i as the endpoint of each interval, \[\huge c_i=\frac{2}{n}i\]this means that we have \[\huge \sum_{i=1}^{n}[2(1+c_i)-5]\Delta x\]

  9. Jonask
    • 3 years ago
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    hence you can steal your function easily from that f(x)=

  10. bmelyk
    • 3 years ago
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    ok

  11. Jonask
    • 3 years ago
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    \[f(x)=2(1+x)-5=2x-3\]@shubhamsrg @ParthKohli

  12. bmelyk
    • 3 years ago
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    and is that the answer?

  13. Jonask
    • 3 years ago
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    \[\int\limits_{-1}^{2}2x-3\]

  14. bmelyk
    • 3 years ago
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    so then i integrate normally?

  15. Jonask
    • 3 years ago
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    yes but did you understand

  16. bmelyk
    • 3 years ago
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    yes i do thank you

  17. Jonask
    • 3 years ago
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    yw

  18. bmelyk
    • 3 years ago
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    is the answer -5?

  19. sirm3d
    • 3 years ago
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    \[\Large \Delta x=\frac{b-a}{n}=\frac{2-(-1)}{n}=\frac{3}{n}\]

  20. bmelyk
    • 3 years ago
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    i subbed in the wrong number

  21. Jonask
    • 3 years ago
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    \[x^2-3x {\huge|}_{-1}^{2}=4-6-1-3=-6\]

  22. bmelyk
    • 3 years ago
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    so you're saying delta c is different than the other perosn?

  23. bmelyk
    • 3 years ago
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    delta x

  24. Jonask
    • 3 years ago
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    i was thinking 3/n but its not even in the equation

  25. bmelyk
    • 3 years ago
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    im sorry i have to go to class but i will be back so please continue to explain further, thank you and chat soon

  26. Jonask
    • 3 years ago
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    are you sure the class is {-1,2} not {0,2}

  27. Jonask
    • 3 years ago
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    so we should write\[\frac{1}{3}\sum_{i=1}^{n}[2(3-2\frac{3i}{n})-5]2\frac{3}{n}\]

  28. Jonask
    • 3 years ago
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    also 5(3) \[f(x)=\frac{2}{3}[2(3-2x)-15]\]

  29. Jonask
    • 3 years ago
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    @amistre64

  30. sirm3d
    • 3 years ago
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    \[\left[2\left(1+2\frac{i}{n}\right)-5\right]\Delta x=\left[2\left(1+\frac{2}{3} \frac{3i}{n}\right)-5\right]\frac{2}{3}\frac{3}{n}\Delta x\] \\[f(x)=\frac{2}3{}\left[2\left(1+\frac{2}{3}x\right)-5\right]\]

  31. amistre64
    • 3 years ago
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    left side is stated as this correct? \[\lim_{n\to0}~\sum_a^b~f(a+i\frac{b-a}{n})\frac{b-a}{n}~;~i=0,1,2,...,b-\frac{b-a}{n}\]

  32. amistre64
    • 3 years ago
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    gpt me i parts mixed around, and put abs in the wrong place, that summation

  33. amistre64
    • 3 years ago
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    hmm interval (-1,2) then b-a= 2--1 = 3 so there seems to be a little confusion unless we are spose to factor within the given summation fo accomodate that

  34. KatClaire
    • 3 years ago
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    @bmelyk @jonask it's [1,3], you typed it in from another question wrong

  35. KatClaire
    • 3 years ago
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    @amistre64 @sirm3d ^

  36. bmelyk
    • 3 years ago
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    yeah i fixed it.

  37. Jonask
    • 3 years ago
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    so is it [-1,2] or [1,3]

  38. sirm3d
    • 3 years ago
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    then it's just \[\Large \int_1^3 (2x-5)\;\mathrm dx\]

  39. Jonask
    • 3 years ago
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    2x-3

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