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Integration by Substitution

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    (i) find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx in terms of X.
    (ii) Find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx for X= 1, 2, 3 and 4.

    2. Relevant equations
    -


    3. The attempt at a solution
    (i) [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex]dx

    -x[itex]^{2}[/itex] = X

    dX/dx=-2x hence -1/2 dX = xdx

    so, -[itex]\frac{1}{2}[/itex] [itex]\int[/itex][itex]^{X}_{0}[/itex] e[itex]^{x}[/itex] dx

    then [itex]\frac{1}{2}[/itex] [1 - e[itex]^{x^{2}}[/itex] ]

    Ok, here is the first problem I have encountered. For my answer, [itex]\frac{1}{2}[/itex] [1 - e[itex]^{x^{2}}[/itex] ], this is wrong according to wolfram and my textbook. the answer should be
    [itex]\frac{1}{2}[/itex] [1 - e[itex]^{-x^{2}}[/itex] ], where there is a minus before the x[itex]^{2}[/itex]. I can not think of how to come to this!! :(

    I assume substituting -x[itex]^{2}[/itex] = X is correct, as that is what i normally do with standard u-substitution problems.

    Ok, second part (ii) So this problem also applies to other problems aswell with negative powers of x.
    Using the answer the textbook got, I input, [itex]\frac{1}{2}[/itex] [1 - e[itex]^{(-1)^{2}}[/itex] ] and get -0.859. The answer should be 0.3161, and to get this I need to remove the brackets from (-1) hence[itex]\frac{1}{2}[/itex] [1 - e[itex]^{-1^{2}}[/itex] ]

    I'm not sure if my brain is playing up and this is a stupid question, or my calculator; but I always thought you place brackets around the negative number when squaring hence -1 x -1 = 1.... rather than -1. This problem also happens with other integration problems I have gone through. Someone help me!
    Thanks for your time!
     
  2. jcsd
  3. Sep 11, 2011 #2

    SteamKing

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    You are jumping ahead of yourself in the choice of substitution. Rather than using the upper limit X of the definite integral, instead choose u = x^2 (where x is the variable of integration).
     
  4. Sep 11, 2011 #3
    Hurray I got it!!! Thank you x1000. But why do I not substitute the minus aswell, so u= -x^2? rather than just u=x^2
    Edit: Checked over it, seems like it only works if I substitute u=-x^2 not u=x^2, .....or not? Confusion! :(
    Do you know why my second part is wrong?
     
    Last edited: Sep 11, 2011
  5. Sep 11, 2011 #4

    SteamKing

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    The substitution u = -x^2 is probably better.

    I don't know why your second part is wrong because nothing of you work is shown.
     
  6. Sep 11, 2011 #5
    Ah sorry, didn't make my workings clear. :p


    (ii) Find [itex]\int[/itex][itex]^{X}_{0}[/itex] xe[itex]^{-x^{2}}[/itex] dx for X= 1, 2, 3 and 4
    I will just do X=1, so I understand how to do the others on my own later.

    [itex]\frac{1}{2}[/itex] [1 - e[itex]^{(-1)^{2}}[/itex] ] and get -0.859.
    The answer should be 0.3161, and to get this I need to remove the brackets from (-1) hence[itex]\frac{1}{2}[/itex] [1 - e[itex]^{-1^{2}}[/itex] ]

    Why do I not include brackets around -1? That would be -1 x 1, hence not (-1)^2 then in my calculator.

    Thanks again :)
     
  7. Sep 11, 2011 #6

    SteamKing

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    In evaluating e^-x^2, for x = 1,2,3,4, the convention is the value of x is squared before application of the minus sign. Similarly, if you were given the polynomial f(x) = -x^2+2x-4 to evaluate at x = -1, you would calculate as follows:

    f(-1) = -[(-1)^2)] + 2*(-1) - 4 = -1-2-4 = -7
     
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