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E^x^3 > 0

  1. Sep 9, 2006 #1

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    http://i5.tinypic.com/4hionb9.jpg

    Please check my answers

    a) I know that e^x^3 > 0 and for all x ∈ [-1, 1], so we have ∫[-1 to 1] e^x^3 dx > 0. But how would you explain in words?

    b) For all x ∈ [0, 1] , 0 ≤ x^2 ≤ x , so e^0 ≤ e^x^2 ≤ e^x
    Therefore
    ∫[0 to 1] e^0 dx ≤ ∫[0 to 1] e^x^2 dx ≤ ∫[0 to 1] e^x dx
    where do i go from here?

    thanks.
     
  2. jcsd
  3. Sep 10, 2006 #2

    HallsofIvy

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    One definition of the definite integral of such a function (positive on the interval) is that it is the "area under the graph". And area is always positive.

    A more rigorous proof is this: f(0)= 1 and the function is continuous so there exist some neighborhod of x= 0, say [itex]-\delta< x< \delta[/itex] such that f(x)> 1/2. We must have [itex]\int_{-\delta}^{\delta}f(x)dx> (2\delta)(1/2)= \delta> 0[/itex]. Since f(x) is never negative, other parts of the integral cannot cancel that: [itex]\int_{-1}^1 e^{x^3}dx> \delta> 0.

    Well, you know that e0= 1 so [itex]\int_0^1 e^0dx= \int_0^1 dx= 1[/itex].
    You also know that [itex]\int e^x dx= e^x[/itex] so [itex]\int_0^1 e^x dx= e^1- e^0= e- 1[/itex]
    What does that tell you about [itex]\int_0^1 e^{x^2}dx[/itex]?
     
  4. Sep 10, 2006 #3

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    then like this?...

    [x] [0 to 1] ≤ ∫[0 to 1] e^x^2 dx ≤ [e^x] [0 to 1]
    1 ≤ ∫[0 to 1] e^x^2 dx ≤ e - 1
    Since 0 < 1 and e - 1 < 3 so we must have 0 < ∫[0 to 1] e^x^2 dx < 3.
     
  5. Sep 10, 2006 #4

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    and also where do all those numbers come from?...8712,08804,8804?
     
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