quick integration help

Discussion in 'Physics & Math' started by Fudge Muffin, May 22, 2013.

  1. Fudge Muffin Fudge Muffin Registered Senior Member

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    148
    Ok, I know you do this by parts but i don't see how...
    how do you integrate (4x)(e^[x squared])

    and apologies, i still have no idea how to write this nicely, in a math algebra-ish way
     
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  3. Fudge Muffin Fudge Muffin Registered Senior Member

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    ah here we go, bascially, what is this???

    \( \Large \int 4xe^{x^2} \)
     
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  5. Tach Banned Banned

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    Use the substitution \(x^2=u\). have fun!
     
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  7. Tach Banned Banned

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    \( \Large \int 4xe^{x^2}=2 e^{x^2}\)


    See previous post for solution
     
  8. rpenner Fully Wired Valued Senior Member

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    4,833
    Integration by parts relies on the product rule of derivatives: \(D[u v] = u D[v] + D[u]v = u v' + u' v\) leads to \(uv =\int u v' dx \, + \, \int u' v dx\) or \(\int u v' dx = uv \, - \, \int u' v dx\)
    If you were to try to solve it by integration by parts you would write (following the LIPET heuristic) \(u = 4x, v' = e^{x^2}\) giving \(u' = 4\) and a complicated odd function for v. \(v(x) = \frac{\sqrt{\pi}}{2} \textrm{erfi}(x)\). Thus we get \(\int \, 4 x e^{x^2} \, dx = 2 \sqrt{\pi} x \textrm{erfi}(x) - \int \, 2 \sqrt{\pi} \textrm{erfi}(x) \, dx = 2 \sqrt{\pi} x \textrm{erfi}(x) - 2 \sqrt{\pi} x \textrm{erfi}(x) + 2 e^{x^2} + C = 2 e^{x^2} + C\) which is horrible if you aren't familiar both with erfi() and its integral.

    http://mathworld.wolfram.com/Erfi.html

    If you knew the definition \(e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}\) you could rely on the linearity of integration which follows from the linearity of derivatives: \(D[a u + b v] = au' + bv'\).
    Thus \(\int \, 4 x e^{x^2} \, dx = \sum_{k=0}^{\infty} \int \, 4 \frac{x^{2k+1}}{k!} \, dx = C + \sum_{k=0}^{\infty} 4 \frac{x^{2k+2}}{2 (k+1) \cdot k!} = C_1 + \sum_{k=0}^{\infty} 2 \frac{x^{2(k+1)}}{(k+1)!} = C_1 + 2 \sum_{k=1}^{\infty} \frac{(x^2)^k}{k!} = C_1 + 2 \left( e^{x^2} - 1 \right) = e^{x^2} + C\)

    But as Tach correctly points out, the simplest way to solve this problem is via u-substitution which follows from the chain rule of derivatives. If \((v \circ u)\) is the composition of functions such that \((v \circ u)(x) = v(u(x))\) then the chain rule is \(D[v \circ u ] = ( v' \circ u ) u'\). So \(D[v(u(x))] = v'(u(x)) \cdot u'(x)\) and if we find a integral that looks like \(\int \, v'(u(x)) \cdot u'(x) \, dx\) we know the answer is \(v(u(x)) + C\).

    Here \(u(x) = x^2, \, u'(x) = 2x \) and by the process of elimination, \(v'(x) = 2 e^x = v(x)\) so \(\int \, 4 x e^{x^2} \, dx = \int \, 2 e^{x^2} \cdot (2x) \, dx = \int \, 2 e^{u(x)} u'(x) \, dx = 2 e^{u(x)} + C = 2 e^{x^2} + C\)
     
  9. rpenner Fully Wired Valued Senior Member

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    4,833
    Sometimes u-substitution can be tricky to find.
    \(\int \, \frac{x^2 -1 }{\left( x^2 + 1 \right)^2} \, dx\)
    \(\int \, \tan x \, dx\)
    \(\int \, \frac{33 x^5 + 24}{3 x^6 + 4x} \, dx\)
     
  10. Tach Banned Banned

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    5,265
    Finding the correct approach is more like black magic, isn't it? This makes integration and solving differential equations so much fun.
     
  11. Tach Banned Banned

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    You are welcome!
     
  12. Fudge Muffin Fudge Muffin Registered Senior Member

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    Yes it is like black magic! I still have no idea how you did that with the substitution, but I found an even better way after staring at it for hours! It turns out if you take the 2 outside, then \( 2x \) becomes part of the integral of \( e^ {x^2} \) which simplifies nicely. Aaaah maths!

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    and thank you tach!!! and rpenner!

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