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E integration

  1. Mar 1, 2004 #1
    how do i do
    [tex] \int_{0}^{\ln 2} e^{-2x} \,dx [/tex]
  2. jcsd
  3. Mar 1, 2004 #2
    Try a substitution of u = -2x. Don't forget to convert the boundaries of integration.

  4. Mar 1, 2004 #3
    Formula : [tex]\int e^{ax}dx= \frac{e^{ax}}{a}+c[/tex]
    Last edited: Mar 1, 2004
  5. Mar 1, 2004 #4
    Figured since himanshu threw out a formula from nowhere, I'd go ahead and derive it. I know I always hated formulas from nowhere.

    [tex]\int_a^b e^{\alpha x}\,dx[/tex]
    make the substitution [itex]u = \alpha x[/itex]. Therefore, [itex]du = \alpha dx[/itex].
    [tex]\int_a^b e^{\alpha x}\,dx = \int_{a'}^{b'} e^{u}\frac{du}{\alpha} = \frac{1}{\alpha} \int_{a'}^{b'} e^{u}\,du [/tex]
    [tex]\frac{1}{\alpha} \int_{a'}^{b'} e^u\,du = \frac{1}{\alpha} e^u \Big|^{b'}_{a'} = \frac{1}{\alpha} e^{\alpha x} \Big|^b_a = \frac{1}{\alpha}(e^{\alpha b} - e^{\alpha a})[/tex]

    Last edited: Mar 1, 2004
  6. Mar 1, 2004 #5
    Just a note: The constant of integration is superfluous in a definite integral, such as this problem. It just cancels itself out at the end anyway.

  7. Mar 2, 2004 #6
    Firstly i wrote it in the form of indefinite integral so from the rul;e i wrote c now u c y i wrote c

    it is not from nowhere
    [tex] \frac{d e^{ax}}{dx}= ae^{ax} [/tex]

    Now rearrange u will get

    [tex]\int e^{ax}dx= \frac{e^{ax}}{a}+c[/tex]
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