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Definite Integration

  1. Jan 24, 2004 #1
    [tex] f(x)=x^2+\int_0^x e^{-t}f(x-t)dt[/tex] ........(I)

    Find f(x)

    Okay what i did:

    For
    [tex]\int_0^x e^{-t}f(x-t)dt[/tex]

    I substituted h=x-t =>dh=-dt
    so [tex]\int_0^x e^{h-x}f(h)dh[/tex]

    Now i differentiated (I)
    so i got
    f'(x)=2x+f(x) after solvin this by integrating factor method i got different results which involve ex

    But i had to prove f(x)=x2+x3/3
     
  2. jcsd
  3. Jan 24, 2004 #2

    Hurkyl

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    The fundamental theorem of calculus says

    [tex]
    \frac{d}{dx} \int_a^x f(h) \, dh = f(x)
    [/tex]

    However, when you differentiate, you don't have something in this form! You have the form

    [tex]
    \frac{d}{dx} \int_a^x f(\mathb{x,} h) \, dh
    [/tex]

    Fortunately, in this case, you don't need the full messy version; you can factor [itex]e^{-x}[/itex] out of the integrand and then differentiate normally.
     
  4. Jan 24, 2004 #3
    Thanks Hurkyl I got corrected

    now i have f'(x)=x2+2x

    which gives me the result

    Still i want to know how would u have approached the pro
     
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