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Integral sqrt(x)*e^-x

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  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate the following integral:
    0 √(x)* e-x dx

    2. Relevant equations
    0 e-x2 dx = (√π)/2

    3. The attempt at a solution
    So far this is what I've done:
    u = x1/2
    du = 1/2 x-1/2

    2 ∫ e-u2 u2 du

    Now, I'm not really sure what to do? Or if what I've done so far is leading me down the right path
     
  2. jcsd
  3. Nov 5, 2016 #2

    andrewkirk

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    Try writing it as ##2\int_0^\infty (e^{-u^2}u)\cdot u\,du## and doing integration by parts. If you can get to where the only remaining integral to do is like ##\int_0^\infty e^{-u^2}\,du## you can use the fact that that integrand is a constant multiplied by the pdf of a standard normal distribution, for which the integral from 0 to ##\infty## is known.
     
  4. Nov 5, 2016 #3
    Okay, so I did (I'm don't remember how to do this very well so bear with me):

    w = u dw = du
    dv = e-u2 u du v = -1/2 e-u2

    (-u/2) e-u2 - ∫ (-1/2) e-u2 du
    -(u/2) e-u2 + (1/2)∫ e-u2 du

    For the left side of the equation, would I plug u back in? If so then the limit as x approaches infinity would make that part 0. For the right side, I'm unsure as to whether I just replace the integral with √π /2. If this is all true, then 2( 0 + (1/2) (√π/2)) and would the final answer be √π/2 ?
     
  5. Nov 5, 2016 #4

    andrewkirk

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    Why have your integration limits disappeared?

    The integration by parts rule for definite integrals is:

    $$\int_a^b f'(t)g(t)\,dt=\left[f(t)g(t)\right]_a^b-\int_a^b f(t)g'(t)\,dt$$
     
  6. Nov 5, 2016 #5

    Okay, so:

    2[[(-u/2) e-u2]0 - ∫0 (-1/2) e-u2 du]
    2[[(-u/2) e-u2]0 + (1/2)∫0 e-u2 du]
    2[(0) + (1/2) (√π/2)] = √π/2
     
  7. Nov 5, 2016 #6

    PeroK

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    Is that ##\frac{\sqrt{\pi}}{2}## or ##\sqrt{\frac{\pi}{2}}##?
     
  8. Nov 5, 2016 #7
    The first one, (√π)/2
     
  9. Nov 5, 2016 #8

    PeroK

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    Looks right.
     
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