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anonymous

  • 5 years ago

Integrate b=0 a=2 x(x-3) dx=??

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{2} x(x-3) dx=?\]

  2. anonymous
    • 5 years ago
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    Multiply it out and integrate term by term. \[= \int\limits_{0}^{2} x^2 - 3x\ dx\]

  3. anonymous
    • 5 years ago
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    (x^3) -(3x^2)/2 and then use the integration limits

  4. anonymous
    • 5 years ago
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    (x^3)/3 -(3x^2)/2 *

  5. anonymous
    • 5 years ago
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    i know the answer is -10/3 but i cant seem to get it. Idk what im missing

  6. anonymous
    • 5 years ago
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    ok so we have (2^3)/3 = 8/3 and 3(2^2)/2 = 3*4/3 = 4 so for the limit 2 we get 8/3 - 4 = and limit 0 will just be 0 for x so it will turn out to be 0. so final answer is 8/3 -4 = 8/3 - 12/3 = -4/3

  7. anonymous
    • 5 years ago
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    woops hold on lol error

  8. anonymous
    • 5 years ago
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    the answer is -10/3 according to the book. I think where you have 4, it should be 6 right? that works out i think

  9. anonymous
    • 5 years ago
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    it shuld be 8/3 - 6 = 8/3 - 18/6 = -10/3

  10. anonymous
    • 5 years ago
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    8/3 - 18/3 = -10/3 *

  11. anonymous
    • 5 years ago
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    thank you for the help... im new to the site... anybody want to give me a quick overview?

  12. anonymous
    • 5 years ago
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    in general, we post doubts and other people try to help...

  13. anonymous
    • 5 years ago
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    i see that... i dont know how i have never heard of this site... it would have been a lifesaver for calculus this semester

  14. anonymous
    • 5 years ago
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    hahaha!

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