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Math Help - Integrating to Find Particular Solution - Question 2

  1. #1
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    Integrating to Find Particular Solution - Question 2

    The answer to the following question is C = -3.75. Please tell me where I went wrong:

    Question: y = integral (squareroot 3x - cuberoot 9x+3x)dx, where x = 3 and y =9

    Let U = 9x + 3x
    du = 9 + 3 dx
    du = 12 dx

    y = integral [(3x)^2] - integral (U)^3

    y = [(3x^3)/3] - [(U^4)/4] + C

    y = x^3 - [(9x+3x)^4]/4 + C

    9 = 3^3 - [(9{3}+3{3})^4]/4 + C

    9 = 27 - [(27+9)^4]/4 + C

    9 = 27 - 1679616/4 + C

    C = 419886 (According to my teacher, the answer is C = -3.75)
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  2. #2
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    Quote Originally Posted by sparky View Post
    The answer to the following question is C = -3.75. Please tell me where I went wrong:

    Question: y = integral (squareroot 3x - cuberoot 9x+3x)dx, where x = 3 and y =9

    Let U = 9x + 3x
    du = 9 + 3 dx
    du = 12 dx

    y = integral [(3x)^2] - integral (U)^3

    y = [(3x^3)/3] - [(U^4)/4] + C

    y = x^3 - [(9x+3x)^4]/4 + C

    9 = 3^3 - [(9{3}+3{3})^4]/4 + C

    9 = 27 - [(27+9)^4]/4 + C

    9 = 27 - 1679616/4 + C

    C = 419886 (According to my teacher, the answer is C = -3.75)
    Your original problem statement is


    Then your first term changes to


    (Edit: You did the same thing with the cube root.)

    Also, I am suspicious about that cube root term. Is there a typo? I can't think of the reason why it wasn't given as 12x in the first place?

    -Dan
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    That's the question: y = integral (square root of 3x minus cube root of (3x + 9x))

    Maybe the 3x + 9x was put there to confuse me. Where did I go wrong in my working?
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    Quote Originally Posted by sparky View Post
    That's the question: y = integral (square root of 3x minus cube root of (3x + 9x))

    Maybe the 3x + 9x was put there to confuse me. Where did I go wrong in my working?
    I have not completed the problem myself, but as I said, you changed into and into .

    Try fixing that and see what happens.

    -Dan
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    Ok, I tried it again and got it wrong again:

    y = integral (square root of 3x minus cuberoot of (9x + 3x))dx

    y = integral of (3x)^2 - integral of (9x)^3 + integral of (3x)^3

    y = [(3x)^3]/3 - [(9x)^4]/4 + [(3x)^4]/4 + C

    y = x^3 - [(12x)^4]/4 + C

    y = x^3 - 3x^4 + C

    9 = 3^3 - (3{3}^4) + C

    9 = 27 - 243 + C

    C = 225

    What am I doing wrong?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sparky View Post
    Ok, I tried it again and got it wrong again:

    y = integral (square root of 3x minus cuberoot of (9x + 3x))dx

    y = integral of (3x)^2 - integral of (9x)^3 + integral of (3x)^3
    What am I doing wrong?
    Your stated problem is


    or should it be


    They are NOT the same!!

    -Dan

    Edit: And




    Edit II:
    I am suspecting that the integral is written incorrectly. Neither form of the problem I listed above gives you the given answer of c = -3.75. The most likely source of error is the 9x + 3x term. I suspect it has been miscopied either by you or the source you got the problem from.
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