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Couple of integrals questions (and with limits)

  1. Feb 8, 2007 #1

    dnt

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    1. The problem statement, all variables and given/known data

    a) integral of e^(x^2) (ie, integral of e to the x squared power)

    b) integral from -1 to 6 of the square root of 3 plus the absolute power of x (sorry i dont know how to make it look nice on these forums...havent learned the code for it yet)

    c) this one is slightly different: express the integral from 1 to 5 of x/(x+1) as a limit


    2. Relevant equations

    n/a

    3. The attempt at a solution


    (a) integral of e^(x^2) (ie, integral of e to the x squared power)

    i know the integral of e^x is itself. but i cant figure out how to adapt to the fact that the x is squared. i tried substitution but that didnt help much.

    (b) integral from -1 to 6 of the square root of 3 plus the absolute power of x

    since the function is always positive, i feel i can just take the integral of the square root of 3 + x (without the absolute power) and plug us -1 and 6 just like normal. is that correct?

    (c) express the integral from 1 to 5 of x/(x+1) as a limit

    always had a little trouble with these...

    i know it will be: the limit as n -> oo (infinity) of the sum from i=1 to n of f(ci)*xi

    (hope you can read that :) )

    the f(ci) represents the heights of the infinite rectangles and the delta xi represents the equal widths of all the rectangles (and you just sum them all up).

    but i always get confused on what to plug into the function itself. in this case x/(x+1). what is ci?

    and also if xi = (b-a)/n (in this case 4/n) how does the integral change if it was from 2 to 6 because xi would again be 4/n? or does it not matter? i would think it should be different since the end points of the function change but xi seems to be the same.

    thanks for any help!
     
  2. jcsd
  3. Feb 8, 2007 #2

    arildno

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    a) Forget it, you won't be able to do that one.
    b) Nope; remember that the absolute value sign is only about the x
    c) You multiply the length of a SMALL INTERVAL with the function value to a chosen point within that interval, and let the number of intervals go to infinity.
     
  4. Feb 8, 2007 #3

    dnt

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    a) thanks

    b) i dont see why that should change anything. do i need to break it into two seperate integrals (one from -1 to 0 and another from 0 to 6)?

    c) still a little confused on how that relates to the function and ci - what do i plug into the function to work with the fact that it represents the height of the infinite rectangles (am i logically saying this correct)?
     
  5. Feb 8, 2007 #4

    arildno

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    b) that would be best.

    c) Divide your interval into (n-1) sub-intervals as follows:
    [tex]1=x_{1}<x_{2}<\cdots<x_{n}=5[/tex]

    The difference [itex]x_{i+1}-x_{i}[/itex] is the length of the i'th sub-interval.

    As your [itex]c_{i}[/itex] value, take for example the midpoint in the i'th interval, [itex]c_{i}=\frac{x_{i+1}+x_{i}}{2}[/itex]
     
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