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From Midterm exam.

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Question 5(10 marks)
    Assume that a function f(x) is continuous on the interval [0,1].Express the following derivatives as formulae in terms of
    [tex]x,f(x),f'(x) and \int_{a}^{x}f(t)dt[/tex]

    [tex]a)\frac{d}{dx}\int_{x}^{0}tf(t)dt[/tex] [tex]b)\frac{d}{dx}\int_{0}^{x}xf(t)dt[/tex]


    [tex]c)\frac{d}{dx}\int_{0}^{x}xf(x)dt[/tex]


    3. The attempt at a solution

    I got 6 marks from 10.Can you help me to find my mistakes?:frown:

    a)[tex]xf(x)[/tex]


    b)[tex]\frac{d}{dx}\int_{0}^{x}xf(t)dt=\int_{0}^{x}f(t)dt+x\frac{d}{dx}\int_{0}^{x}f

    (t)dt=\int_{0}^{x}f(t)dt+xf(x)[/tex]

    c)[tex]\frac{d}{dx}\int_{0}^{x}xf(x)dt=\int_{0}^{x}f(x)dt+f'(x)x\int_{0}^{x}dt+xf

    (x)\frac{d}{dx}\int_{0}^{x}dt=f(x)x+f'(x)x^2+xf(x)=2xf(x)+x^2f'(x)[/tex]
     
  2. jcsd
  3. Oct 21, 2007 #2

    quasar987

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    c) is right

    for a), the FTC says that

    [tex]\frac{d}{dx}\int_{0}^{x}tf(t)dt=xf(x)[/tex]

    so you forgot to invert the bounds of the integrals first, which would induce a minus sign in the answer.

    For b), this is an application of Leibniz rule of differentiation under the integral sign where the integrand xf(t) is considered as a function of two variables g(x,t)=xf(t).
     
  4. Oct 21, 2007 #3
    then [tex]\frac{d}{dx}\int_{0}^{x}g(x,t)dt[/tex].


    Right?
     
  5. Oct 21, 2007 #4

    quasar987

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    what's your question?
     
  6. Oct 21, 2007 #5
    [tex]\frac{d}{dx}\int_{0}^{x}g(x,t)dt=\frac{d}{dx}\int_{0}^{x}xf(x)dt[/tex]
     
  7. Oct 21, 2007 #6

    quasar987

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    No,

    [tex]\frac{d}{dx}\int_{0}^{b(x)}g(x,t)dt = \int_{0}^{b(x)}\frac{\partial}{\partial x}g(x,t)dt+b'(x)g(x,b(x))[/tex]

    So

    [tex]\frac{d}{dx}\int_{0}^{x}xf(t)dt = \int_{0}^{x}f(t)dt+xf(x)[/tex]
     
    Last edited: Oct 21, 2007
  8. Oct 21, 2007 #7
    Very nice.I've haven't seen this method.
     
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