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Integral of -e^(-x)

  1. Sep 21, 2004 #1
    Hi guys, I am looking in my calculus book, and it tells me in the example problem that the Integral of e^(-x)dx = -e^(-x)

    I dont see how u get this answer. I know the integral of e^x = e^x but how does that negative sign come out to the left? Can someone explain this step by step, thanks.
  2. jcsd
  3. Sep 21, 2004 #2
    If you try differentiating -e-x, it comes to (-1)-e-x which is e-x. So it makes sense!
  4. Sep 21, 2004 #3
    Yes, but how do u get the integral without knowing thats the answer?
  5. Sep 21, 2004 #4
    Oh crap nevermind, its easy, just use u substitution u = -x
  6. Sep 21, 2004 #5
    This is a theorem.

    The integral of e^u du = e^u * (du/dx).
  7. Mar 22, 2009 #6
    If you have e^(-x), you must divide e^(-x) by (-1), cuz (-1) is the coefficient of (-x)

    that's the way you integrate

    when you have e^(-3x) you must divide e^(-3x) by (-3)... so on...
  8. Mar 23, 2009 #7


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    or more simply, divide by the derivative of the power.

    [tex]\int_a^b{e^{f(x)}}=\left [\frac{e^{f(x)}}{f'(x)} + C \right ]^b_a[/tex]
  9. Mar 23, 2009 #8
    I don't think that your last bit is true in general... I think that
    [tex]\int_a^b{ f'(x) e^{f(x)}}dx=\left [e^{f(x)} + C \right ]^b_a[/tex]
    is what you mean and you have divided through. This is only correct when the f'(x) is constant... you might have known this, I just found it confusing the way you stated. Also for definite integrals you can omit the constant.

    For example
    [tex]\int_a^b{e^{-x^{2}}}=\left [\frac{\sqrt{\pi}}{2}erf(x) \right ]^b_a[/tex]
  10. Mar 23, 2009 #9


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    That is completely wrong. In order to use "substitution" with u= f(x), f'(x), unless it is a constant, must already be in the integral.
  11. Mar 23, 2009 #10
    You have been given:
    [tex]\int e^u du = e^u + C[/tex]
    where C is an undetermined constant of integration.
    You have come upon the integral:
    [tex]\int f(g(x)) dx[/tex]
    where you know the indefinite integral of f(u), but not f(g(x)). In your case f(u) = eu and g(x) = -x. If we go back to derivatives, we may remember a related rule called the chain rule:
    (f(g(x))' = f'(g(x))*g'(x)
    In other words:
    [tex]\int f'(g(x))*g'(x) dx = f(g(x)) + C[/tex]
    In some texts, they will use shorthand Leibnitz notation: Let u = g(x). Then du = g'(x) dx. The above integral is then written:
    [tex]\int f'(u) du = f(u) + C[/tex]
    This is usually called u-substitution. Note that you have to have both f'(g(x)) = f'(u) and du = g'(x) dx in the integrand. Sometimes you will have to multiply the integrand by a creative version of 1 in order to make this happen. In your example, let f'(u) = eu since we already know how to integrate that and of course u = -x. Then du = -1 dx. Then we need our integrand to be f'(u) du = -e-x dx. Unfortunately, our integrand is actually -(e^-x dx) = -f'(u) du. Luckily, the -1 can be factored out of the integrand as it is a constant, so we have
    [tex]-\int f'(u) du = - f(u) + C = -e^u + C[/tex]
    Last edited: Mar 23, 2009
  12. Mar 23, 2009 #11
    Here is a simple rule you can use:
    \int f(ax + b) \, dx = \frac{F(ax + b)}{a} + C

    where F(x) is a primitive function of f(x). We can use this like so:
    \int e^{-x} \, dx = \frac{e^{-x}}{-1} + C
  13. Mar 24, 2009 #12


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    Thanks for correcting me. It seems highschool doesn't expect any more than the elementary integrals of ef(x), f(x) being a linear function.
    Hallsofivy, I'm not exactly sure what you're saying. This result has been given to us in class.

    Maybe, [tex]\int f'(x)e^{f(x)}=\frac{f'(x)e^{f(x)}}{f'(x)}+C=e^{f(x)}+C[/tex]
  14. Mar 24, 2009 #13
    Halls is correct; your previous result makes no sense unless f(x) is linear, in which case f'(x) is just a constant, from which you can derive the result yourself from the chain rule. The far left and far right hand sides of the above quoted integral are correct, and is a straightforward result of the chain rule; your middle step is unjustified.
  15. Jul 7, 2010 #14
    listen to this guy k.. the substitution method is sooooo not necessary in this case.. it just complicates something that is actually very very simple.. above is the general method of integrating exponentials.. it might be easier if u understand integration as the opposite of differentiation.. when differentiating exponentials you multiply by the derivative of function x.. and integrating is the opposite so you divide by the derivative of function x.. anyway good luck :)
  16. Jul 7, 2010 #15


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    Yes sorry, I was only thinking of functions f(x) that were linear y=ax+b.
  17. Jul 7, 2010 #16


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    And after reading the rest of the posts, it seems like my restriction has been mentioned multiple times.

    The only quarrel they had was that the result I gave is only correct if f(x)=ax+b, [itex]a\neq[/itex]0. It was my mistake. We should instead stick to the proper formula :smile:
    And yes I agree with you. In this problem substitution should be avoided and that's what I try to get students to do - think of integrals as back-differentiation - when solving something so simple.
  18. Sep 6, 2010 #17

    The integral of 2x.e^(x^2) is e^(x^2), isn't it?

    Surely the integral of f'(x).e^f(x) is always e^f(x), regardless of whether f'(x) is a constant?
  19. Sep 6, 2010 #18


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    Yes, but what I did was rather than saying [tex]\int{f'(x)e^{f(x)}dx}=e^{f(x)}+c[/tex] I mistakenly wrote [tex]\int{e^{f(x)}dx}=\frac{e^{f(x)}}{f'(x)}+c[/tex] which is only true for f(x)=ax+b since then f'(x) is a constant.
  20. Sep 6, 2010 #19
    u = -x
    du = -1dx
    dx = -du

    so you subsitute e-x for eu and then substitute dx for -du, pull out the negative infront of the integral and you get -eu + C, subsituting back in and get:

    -e-x + C
  21. Sep 6, 2010 #20


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    A well kept secret - you practically can't get any integral without knowing that's the answer.
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