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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Question is....


    Use Integration by parts to show that the

    integral f(x)dx = xf(x) - integral xf'(x)dx

    What do I need to make my U? DV? I have tried all kinds of combinations of the substitution method and just can't figure it out. HELP!!!! TIA John
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  2. #2
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    We will assume that f(x) is a differenciable function on some open interval 'I'.
    Therefore, it is countinous.
    Therefore, there exists a function F(x) definied on open interval 'I' such that, F'(x)=f(x).
    Now we can proceede.
    ---
    You are given,
    Integral f(x)dx,
    You can express it as,
    Intergral (1)f(x)dx,
    Let u'=x and v=f(x)
    Then,
    u=1 and v'=f'(x)
    By integration by parts,
    f(x)-Integral xf'(x)dx
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    We will assume that f(x) is a differenciable function on some open interval 'I'.
    Therefore, it is countinous.
    Therefore, there exists a function F(x) definied on open interval 'I' such that, F'(x)=f(x).
    Now we can proceede.
    ---
    You are given,
    Integral f(x)dx,
    You can express it as,
    Intergral (1)f(x)dx,
    Let u'=x and v=f(x)
    Then,
    u=1 and v'=f'(x)

    if u=1, u'=0 or if u'=x u=(x^2)/2

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    if u=1, u'=0 or if u'=x u=(x^2)/2

    RonL
    My mistake,
    u'=1 and v=f(x)
    Then,
    u=x and v'=f'(x)

    You have to forgive me, I do intergrations in a formal way.
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  5. #5
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    Quote Originally Posted by TreeMoney View Post
    Question is....


    Use Integration by parts to show that the

    integral f(x)dx = xf(x) - integral xf'(x)dx

    What do I need to make my U? DV? I have tried all kinds of combinations of the substitution method and just can't figure it out. HELP!!!! TIA John
    Consider:

    int x f'(x) dx,

    let u=x and v'=f', then:

    int x f'(x) dx = uv - int u' v dx=x f(x) - int f(x) dx.

    Now rearrange:

    int f(x) dx = x f(x) - int x f'(x) dx

    RonL
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  6. #6
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    Part B :o(

    PerfectHacker, I made the same mistake you made originally!!! Thanks again to everyone for your help!!!

    There is another part to this question... It says.

    If f and g are inverse functions and f' is continuous, prove that

    the integral(from a to b) f(x)dx = bf(b) - af(a) - integral(from f(a) to f(b)) g(y)dy

    Then there is a hint, it says to use part (a) and make the substitution y = f(x)

    I tried to figure this out, but kept getting stuck, To be honest I probably wasn't even close to having the right answer. PLEASE HELP!!!
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by TreeMoney View Post
    There is another part to this question... It says.

    If f and g are inverse functions and f' is continuous, prove that

    the integral(from a to b) f(x)dx = bf(b) - af(a) - integral(from f(a) to f(b)) g(y)dy
    The definite integral that we get from what we have already found is:

    int(a,b) f(x) dx = b f(b) - a f(a) -int(a,b) x f'(x) dx

    Now let y=f(x), then dy = f'(x) dx, x=g(y), and the limits of integration
    become y1=f(a) and y2=f(b), so putting all of this into the integral on the
    right hand side we have:

    int(a,b) x f'(x) dx = int(f(a),f(b)) g(y) dy,

    so:

    int(a,b) f(x) dx = b f(b) - a f(a) - int(f(a),f(b)) g(y) dy

    which is what was to be prooved.

    RonL
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