integration

Discussion in 'Math' started by slimnick, Jan 31, 2008.

  1. slimnick

    Thread Starter Member

    Nov 25, 2007
    21
    0
    hello, i am currently in my second semester in math at my college and there are integral problems given. is there any hints or help that you guys can give to help me out to work the porblem. maybe which froms of intergation to use or what is the best way to start. any help wioll be appreciated.how to start.
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    Would it make it easier if I said:

    \sqrt{x}=x^{0.5}

    Dave
     
  3. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    I'm not sure how much of a hint this is, Dave.

    The integral provided by the OP is a bit tricky, IMO. To carry it out I used two substitutions to get an improper rational function, carried out the division to get a polynomial and a proper rational function, and finally integration by partial fractions. The two substitutions could be replaced by one judicious substitution, of course, and the partial fractions bit could be done be looking in a table (which uses partial fractions under the hood).

    Here's more of a hint using one substitution: Let v = (1 + e^x)^0.5. Then v^2 = 1 + e^x, so 2v*dv = e^x * dx.

    This substitution should produce [Integral]2v^2 /(v^2 -1) dv,
    for which the division can be done to produce [Integral] (1 + 1/(v^2 - 1)) dv.

    [Integral] 1 dv is easy.
    [Integral] 1/(v^2 - 1) dv can be gotten by partial integration or by looking in a table.

    After you have carried out the integration above, undo the subsitution to get the integral of the original expression.
    Mark
     
  4. slimnick

    Thread Starter Member

    Nov 25, 2007
    21
    0
    i dont understand this all the way. maybe it slipped my mind at the moment, but how could 2v*dv be used as this, which tenchique is this. im still alittle confused. ill try to follow through for now but i think i need some clarity
     
  5. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    With this substitution, your original integral -- [Integral](1 + e^x)^0.5 dx
    becomes:
    [Integral] (2v^2)/(v^2 - 1) dv

    To do the substitution, you need to be able to take differentials, which you probably learned in the first semester.

    If v = (1 + e^x)^0.5, then v^2 = 1 + e^x
    (I squared both sides to make it easier to get differentials.)
    So 2v * dv = e^x * dx

    We're going to have to use this substitution to replace all x's and dx's in the original integral, so we might as well solve the last equation above for dx.

    Doing that, we get dx = 2v/(e^x) * dv
    but since v^2 = 1 + e^x, then e^x = v^2 - 1

    This means that dx = 2v/(v^2 - 1) * dv

    Replacing x's and dx's with v's and dv's in the original integral, we have
    [Integral](1 + e^x)^0.5 dx = [Integral] v * (2v * v/(v^2 - 1)) * dv

    2v^2
    IOW, the integrand is -------
    v^2 - 1

    This is an improper rational function because the degree of the polynomial in the numerator is 2, which is equal to the degree of the polynomial in the denominator.

    If you do the division, you get 1 + 1/(v^2 - 1) as the new integrand.

    All you have to do now is evaluate [Integral] ( 1 + 1/(v^2 - 1)) * dv, which is reasonably straightforward, and finally undo the substitution to get the answer to the original integration problem.

    Hope that helps...
    Mark
     
  6. slimnick

    Thread Starter Member

    Nov 25, 2007
    21
    0
    yes i have dont differential in this first smester, thanx alot for helpin, your ideas were very helpful:)
     
  7. aliashar86

    Active Member

    Nov 23, 2006
    71
    0
    u first need to build your concepts clear with "derivatives" then come up with integration because both are opposite to each other
     
  8. aliashar86

    Active Member

    Nov 23, 2006
    71
    0
    like for example

    the derivative of "x" is 1 while the integral of "1" (which is the answer) is "x" so both are converse to each other.
     
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