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Math Help - Series integration

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    Series integration

    We have to calculate Integrate[e^(-x^2),{x,0,0.1}] using series theory. Any help?
    Thanks!
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  2. #2
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    Quote Originally Posted by jmytil View Post
    We have to calculate Integrate[e^(-x^2),{x,0,0.1}] using series theory. Any help?
    Thanks!
    <br />
\int_0^{0.1} e^{-x^2}dx<br />

    We will do this by expanding the integrand as a power series and then integrating term by term.

    <br />
e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}<br />

    so:

    <br />
e^{-x^2}=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}<br />

    Hence:

    <br />
\int_0^{0.1} e^{-x^2}dx=\int_0^{0.1} \left[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}\right] dx<br />

    Now we will interchange the order of the integration and summation, which
    is valid in this case but I will leave the justification for you to provide:

    <br />
\int_0^{0.1} e^{-x^2}dx= \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!} \int_0^{0.1}x^{2n} dx \right]<br />

    so:

    <br />
\int_0^{0.1} e^{-x^2}dx= \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!(2n+1)} (0.1)^{2n+1} \right]<br />

    Summing the first two terms on the right gives an estimate for the integral of 0.0996666667
    compared to a fairly accurate numerical integration which gives: 0.0996676643, and the
    sum of the first five terms of the series give: 0.0996676643.

    RonL
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