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Thread: help needed

  1. #1

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    help needed

    Hi, i am new to this forum and could do with some help on a question on intergration.

    3
    / (3x(power of 5)-2x-3)dx
    2

    sorry about the format, any help appreciated.

  2. #2
    Member Thomas154321's Avatar
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    Re: help needed

    ∫ 3x^5 -2x -3 dx

    = (x^6)/2 –x2 -3x.

    So with limits of 2 and 3, sub x=2 and take that from what you get subbing x=3.
    = (346.5) - (22)
    =324.5

  3. #3
    Smitten by reality Harsh Gupta's Avatar
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    Re: help needed

    ok i am denoting the integration sign with limits like 3/2
    Code:
    3/2 (3x^5 - 2x - 3) dx
    
    = 3/2 (3x^5)dx - 3/2 (2x)dx - 3/2 (3)dx
    
    = 3/2 (3(x^6)/6) - 3/2 (2(x^2)/2) - 3/2 (3x)
    
    = ((3^6)/2 - (2^6)/2) - (3^2 - 2^2) - (3*3 - 3*2)
    
    = (364.5 - 32) - ( 9 - 4 ) - (9 - 6)
    
    = 332.5 - 5 - 3
    
    = 324.5
    sorry, but i think i am wrong. please someone correct me
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  4. #4
    Smitten by reality Harsh Gupta's Avatar
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    Re: help needed

    hey thomas, how you made the integral sign??
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  5. #5
    Member Thomas154321's Avatar
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    Re: help needed

    I copied it from word.

  6. #6
    Smitten by reality Harsh Gupta's Avatar
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    Re: help needed

    Quote Originally Posted by Thomas154321
    I copied it from word.
    :slaps forehead:
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  7. #7

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    Re: help needed

    thanks guys kinda understand it, but i seem to be getting a different answer, here is a simpler one, maybe i will be able to follow your workings better. Sorry if i seem a bit slow, new starter you see,

    3
    / x dx
    2

    thanks.

  8. #8
    Member Thomas154321's Avatar
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    Re: help needed

    To integrate the sum of terms, take each part separately. So first, take the 3x^5.

    1) Add 1 to the power = 3x^6.

    2) Divide by the new power = (3x^6)/6 = (x^6)/2.

    3) ∫(-2x) dx = (-2x^2)/2 = -x^2.

    4) ∫(-3) dx = -3x.

    So, add them up to get (x^6)/2 -x^2 -3x

  9. #9

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    Re: help needed

    Hi, think i have grasped it,
    would i be right in saying that,

    3
    / x dx would be, (3^2/2) - (2^2/2) = 9/2 - 4/2 = 4.5-2 =2.5.
    2

    thanks astro.

  10. #10
    Member Thomas154321's Avatar
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    Re: help needed

    Yes. That is correct.

  11. #11

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    Re: help needed

    thanks for all your help thomas and harsh gupta, could please re assure me on this last question, thanks,

    2
    / (4x^2 - 5x + 2) dx
    -1

    (4x^3/3 - 5x^2/2 +2x)

    (x^3/0.75 - x^2/2.5 + 2x)

    (2^3/0.75 - -1^3/0.75) - (2^2/2.5 - -1^2/2.5) + (2*2 - 2*-1)

    (10.7 - -1.3 - 1.6 - 0.4 + 6) = 16

  12. #12
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    Re: help needed

    2
    ∫ (4x^2 - 5x + 2) dx =
    -1

    [(4/3)x^3 - (5/2)x^2 + 2x] from 2 to -1 =

    ((32/3)-10+4) - (-(4/3)-(5/2)-2) =
    (14/3)-(-105/18) = 10.5.

    You divided the middle term by 2.5 instead of multiplied. You did the basic integration correctly though, which is the hard part.

  13. #13

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    Re: help needed

    sorry thomas could you simplify your workings for me, sorry if i seem a bit slow. thanks astro

  14. #14
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    Re: help needed

    Ok, to integrate 4x^2 - 5x + 2, the 4x^2 goes to (4x^3)/3, the -5x goes to (-5x^2)/2, and the +2 goes to +2x.

    Add these 3 to get (4x^3)/3 (-5x^2)/2 +2x.

    Replace x by 2 and get a number, and then replace x by (-1) to get a second number. Take the second away from the first to get the answer.

  15. #15

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    Re: help needed

    thanks for all your help thomas, its much appreciated. Its finally clicked now

    regards astro

  16. #16
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    Re: help needed

    Pleasure.

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