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Integration of 2^{-x}

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Waht is the correct integral of

    2. Relevant equations


    2^{-x}

    3. The attempt at a solution

    Is -{1/(x-1)}*2^{1-x} correct or do I have to apply some substitution?


    Thanks for your answers!
     
  2. jcsd
  3. Dec 12, 2012 #2

    Simon Bridge

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    Welcome to PF;
    Hint: You can check your solution by differentiating it.
    If it comes back with what you started with, you got it right.

    It also helps if you state your reasoning - it looks like you've tried the rule for integrating powers of the variable.
    Here, the variable is the power, does that make sense?
     
  4. Dec 12, 2012 #3
    Try this:

    2 = e ^(ln 2)

    ln2 is defined as "the power you have to raise e to to get the number 2"

    So the number 2 is equal to e raised to "the power you have to raise e to to get the number 2"
     
  5. Dec 12, 2012 #4

    HallsofIvy

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    You need to learn very quickly and very thoroughly that the derivative formula [itex]d(x^n)/dx= nx^{n-1}[/itex] and the corresponding integral formula [itex]\int x^n dx= 1/(n+1) x^{n+1}+ C[/itex] only hold when the variable, x, is the base and the exponent is a constant. The situation in which the base is a constant and the exponent is x, is completely different. Both derivative and integral, for example, of [itex]e^x[/itex] is just [itex]e^x[/itex] itself (plus "C" for the integral, of course).

    To differentiate something like [itex]e^{f(x)}[/itex], you need to use the chain rule:
    Let u= f(x) so that [itex]e^{f(x)}= e^u[/itex]. Then [itex]d(e^u)/dx= e^u du/dx[/itex] or [itex]d(e^{f(x)}/dx= e^{f(x)}df/dx[/itex]. The integral is harder- we cannot use "substitution", which is essentially the "inverse" of the chain rule, unless we already have the "df/dx= f'(x)" in the integral: [itex]\int e^{f(x)}f'(x)dx= e^{f(x)}+ C[/itex].

    In the special case that f(x) is "linear", that is [itex]f(x)= ax+ b[/itex] (like your example here, a= -1, b= 0) then f'(x)= a, a constant, and we can take a constant in and out of an integral at will. To integrate [itex]\int e^{-x}dx[/itex], let u= -x so that du= -dx and now there are two ways of thinking:
    1) Multiply by (-1)(-1)= 1, taking one -1 inside the integral: [itex]\int e^{-x}dx= -\int -e^{-x}dx[/itex] and now let u= -x so that du= -dx and [itex]-\int -e^{-x}dx= -\int e^{-x}(-dx)= -\int e^u du[/itex][itex]- e^u+ C= -e^{-x}+ C[/itex].

    2) Start of by letting u= -x so that du= -dx and -du= dx so that [itex]\int e^{-x}dx= \int e^u(-du)= -\int e^u du= e^u+ C= e^{-x}+ C[/itex]

     
  6. Dec 12, 2012 #5

    Ray Vickson

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    You should never ask, in this type of question, whether a proposed solution is correct; you should always check by taking the derivative to see whether you get back to the original function. Try it!
     
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