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Math Help - integral

  1. #1
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    integral

    Hi i need some help please,,
    Im going integration,
    i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
    But then i tried to do the others but they arent working out..
    s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S

    But then there are some with brackets in and i dont know how to go about those help please.
    s^6 ^5 (2x+1)dx ...
    Thank you
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by Chez_ View Post
    Hi i need some help please,,
    Im going integration,
    i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
    But then i tried to do the others but they arent working out..
    s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S

    But then there are some with brackets in and i dont know how to go about those help please.
    s^6 ^5 (2x+1)dx ...
    Thank you
    You need to work on your notation.

    play around with latex in this thread.

    try first writing
    [PHP] \int_{0}^{3} 2x~dx[/PHP]
    It should look like this:
    \int_{0}^{3} 2x~dx

    The \int means "integral" the _{0} means to put a zero at the bottom, the ^{3} means to put a 3 at the top. the ~ adds a space.

    Anyway
    \int_0^2 2x~dx ~~~~=~~~~ [x^2]_0^3 ~~~~=~~~~(3)^2-(0)^2~~~~=~~~~3^2~~~~=~~~~9




    And for the one with brackets, just split the integral up like I did here, or you can keep them in the same group and just integrate them independently
    \int_5^6 (2x+1)~dx

    =\int_5^6 2x~dx+\int_5^6 1~dx

    =[x^2]_5^6~~+~~[x]_5^6

    =[6^2-5^2]~~+~~[6-5]

    =6^2-5^2+6-5

    =36-25+6-5

    =12
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by Chez_ View Post
    Hi i need some help please,,
    Im going integration,
    i have done the first question = (ment to b a math symbol like a f with no stick)S^2 and^1 2xdx which indifferentated =2^2-1^2 = 3 and got the right answer.
    But then i tried to do the others but they arent working out..
    s^3 & ^0 2xdx = x^2, then 2^3-0^2 which =8 but the answe is 9..:S ..... wow!

    But then there are some with brackets in and i dont know how to go about those help please.
    s^6 ^5 (2x+1)dx ...
    Thank you
    To calculate

    \int_5^6 (2x+1)dx you calculate the integral of each summand:

    \int_5^6 (2x+1)dx = \left. x^2 + x \right|_5^6

    Now plug in the values and calculate the difference. You should get 12.
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by earboth View Post
    To calculate

    \int_5^6 (2x+1)dx you calculate the integral of each summand:

    \int_5^6 (2x+1)dx = \left. x^2 + x \right|_5^6

    Now plug in the values and calculate the difference. You should get 12.
    Breathing down my neck!
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  5. #5
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    Hey guys thanks for all your help i have now done the 9 on the exercise but i just have these three tricky ones to do, i have tried a few things but havnt got the answer..
    S^3^0 2xdx

    S^5^1 xdx
    and finally

    S^4^-5 (x^3+3x)dx

    Thank you
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  6. #6
    Super Member angel.white's Avatar
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    Write this one:
    S^3^0 2xdx

    Like this: \int_{0}^{3} 2x~dx
    Highlight it, and click it should be located near the other options like bold, center, hyperlink, etc. If you move your mouse over a math equation, you will see what the person wrote in order to get it to work, so move your mouse over this:
    \int_{0}^{3} 2x~dx

    a little box should pop up and show you what I just said.

    -----------------

    \int^{4}_{-5} (x^3+3x)~dx

    =\int^{4}_{-5} x^3~dx~~+~~\int_{-5}^{4}3x~dx

    Can you continue from here?
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