How do you integrate this problem?

int over [0.23847, 1] of 5 [ sqrt {x} - e^(-3x) ] ^2 dx

I'm getting a bit confused.

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  • 1 decade ago
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    I think this integral has no analytical solution

    ∫ 5 · [√x - e^(-3·x) ]² dx

    = 5·∫ x - 2·√x·e^(-3·x) + e^(-6·x) dx

    = 5·[ ∫ x dx - ∫ 2·√x·e^(-3·x) dx + ∫ e^(-6·x) dx ]

    1st and 3rd integral are simple:

    ∫ x dx = (1/2)·x²

    ∫ e^(-6·x) dx = -6·e^(-6·x) dx

    The 2nd integral is non-analytical.

    You can express in terms of error function.

    Error function is defined as:

    erf(x) = (2/√π) · ∫|0→x| e^(-t²) dt

    so substitute

    z = √(3·x) <=> z² = 3·x => dx = (2/3)·z dz

    ∫ 2·√x·e^(-3·x) dx

    = ∫ 2 · (z/√3) · e^(-z²) · (2/3)·z dz

    = (4/9)·√3 · ∫ z²· e^(-z²) dz

    Integrate by parts

    (remember that d(e^(-z²))/dz = -2·x·e^(-z²)

    = (2/9)·√3 · ∫ z · ( 2·z·e^(-z²) ) dz

    = (2/9)·√3 · [ z·(-e^(-z²)) - ∫ 1· (-e^(-z²) dz

    = (2/9)·√3 · [ ∫ e^(-z²) dz - z·e^(-z²)) ]

    = (2/9)·√3 · [ (√π/2)·erf(z) - z·e^(-z²)) ]

    = (1/9) · [ (√(3·π)·erf(z) - √3 ·z·e^(-z²)) ]

    = (1/9) · [ (√(3·π)·erf(√(3·x)) - 6·x·e^(-3·x)) ]

    Therefore

    ∫ 5 · [√x - e^(-3·x) ]² dx

    = (5/2)·x² - (5/9)·[ (√(3·π)·erf(√(3·x)) - 6·x·e^(-3·x)) ] - 30·e^(-3·x)

    The rest is cake, just insert your limits.

    To calculate the values of erf(3·x) you can either use tabulated values . e.g:

    http://www.math.hkbu.edu.hk/support/aands/toc.htm

    or use online calculator like:

    http://functions.wolfram.com/webMathematica/Functi...

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  • 1 decade ago

    answer is 1.554354405333

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