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anonymous

  • 5 years ago

Use the midpoint rule: n=3, integral from -5 to 0, (-2x-6x^2)dx

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  1. anonymous
    • 5 years ago
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    *grunt* having to do summation finally, you do it

  2. anonymous
    • 5 years ago
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    doing it lol

  3. anonymous
    • 5 years ago
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    -5 to 0 is 5 so with n=3 your dx will be 5/3. This means that your first midpoint will be at (-5/3)/2 or -5/6 and all subsequent ones will be dx multiples away. So using this you can then calculate the rieman sum 5/3[{-2(-5/6)-6(-5/6)^2}+{-2(-15/6)-6(-15/6)^2}+{-2(-25/6)-6(-25/6)^2}]. You can do the math on that one i dont have a calculator near me. explanation to follow.

  4. anonymous
    • 5 years ago
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    THANK YOU SOOOOOOO MUCH!!!!!!

  5. anonymous
    • 5 years ago
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    All i did was apply the rule for the mid point rule which is Q[F(a+Q)+F(a+2Q)+F(a+3Q)...all the way up to F(a+NQ) where N is the amount of divisions they want, Q is the range of the integral divided by N (in your case 5/3) and a is the first midpoint that you use (in your case it was -5/6 because it was half the distance of the base but this is not for all cases because you could have the same base with a range of -10 to -5 but the first point would be -105/6) http://en.wikipedia.org/wiki/Riemann_sum#Middle_sum

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