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Math Help - definte integrals

  1. #1
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    definte integrals

    hi

    can some one please check this for me

    need to evaluate integral x(7-3x^2) for x=6 to x=4

    i did this 1/2(7-3x^2)(2x)dx

    1/2 *1/2(7-3x^2)^2+c

    1/4(7-3x^2)^2+c

    then for x=6 to x=4

    1/4(7-3(6)^2-1/4(7-3(4)^2

    am i on the right track ?

    wayne
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  2. #2
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    Some of your algebra is a bit worrying - check carefully.

    Your integral is easy (no chain or product rule to work backwards through, just the power rule) if you start with

    x(7 - 3x^2) = 7x - 3x^3

    Then integrate each term separately.
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  3. #3
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    Don't integrate - balloontegrate!

    Balloon Calculus Forum
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  4. #4
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    you came to my help again thanks

    i get x(7-3x^2)dx to be

    7x-3x^3
    integrate= 7/2(x^2)-3/4(x^4)

    if this is wrong then i think i will change my math course to history course.
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  5. #5
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    That's fine, and brackets in order... hang on in there!

    (Plug in the limits...)
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  6. #6
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    plugged in limits and got answer to be 31.

    thanks for you patients and help

    wayne
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  7. #7
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    You are plugging the limits into your (correct) integral expression, are you?

    \int_4^6 (7x - 3x^3) dx = \large{[}\frac{7}{2}x^2 - \frac{3}{4}x^4\large{]}_4^6

    = [\frac{7}{2}6^2 - \frac{3}{4}6^4] - [\frac{7}{2}4^2 - \frac{3}{4}4^4]

    = ...
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  8. #8
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    thanks for checking that now have -1094
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  9. #9
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    Brackets! (Signs!)
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  10. #10
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    eureka i have it -710

    i get (7/2(6)^2)-(3/4(6)^4)= -846
    and (7/2(4)^2)-(3/4(4)^4)= -136

    -846--136= -710

    if this is not correct i will eat my calculator.
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  11. #11
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    Good. Then reflect that there are no negative areas...
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