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    Hey, got a really odd answer for this question... could someone please confirm it?! integral of (x-1)(x+1)^3 dx with limits 2->0 , integration by parts. Sorry none of the symbols work on this comp. I got -236.25 - somehow dont think that's right!! Thanks
    franks xxx
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    If that was (x+1)^3 multiplied by (x-1), then the answer should be 42/5, which I did by multiplying out the brackets, but what values of u and v did you use?
    Unless i misunderstood the question, it is awkward without symbols.
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    To check, you can make the integral easy by using the sub y = x+1
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    (Original post by franks)
    Hey, got a really odd answer for this question... could someone please confirm it?! integral of (x-1)(x+1)^3 dx with limits 2->0 , integration by parts. Sorry none of the symbols work on this comp. I got -236.25 - somehow dont think that's right!! Thanks
    franks xxx
    use integration by substitution:

    let u = x + 1
    du/dx = 1 thus du = dx

    x + 1 = u - 2

    then you get

    ∫u³(u - 2)du, with limits 3, 1

    which is just ∫u^4 -2u³ du

    simple!
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    (Original post by franks)
    Hey, got a really odd answer for this question... could someone please confirm it?! integral of (x-1)(x+1)^3 dx with limits 2->0 , integration by parts. Sorry none of the symbols work on this comp. I got -236.25 - somehow dont think that's right!! Thanks
    franks xxx
    u = (x-1), dv/dx = (x+1)³
    du/dx = 1, v = ¼(x+1)4

    ∫(x-1)(x+1)³dx
    = ¼(x-1)(x+1)4 - ¼∫(x+1)4dx
    = ¼(x-1)(x+1)4 - (x+1)5/20

    Definite integral between 2 and 0
    = ¼(2-1)(2+1)4 - (2+1)5/20 - (¼(0-1)(0+1)4 - (0+1)5/20
    = 81/4 - 243/20 + 1/4 + 1/20
    = 42/5
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    maaaan Im so STUPID !!!! I just assumed that when you inserted 0 into the equation, it came to 0 - thanks so much for shoving me back in the right direction !!
    franks xxx
 
 
 
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