We have to calculate Integrate[e^(-x^2),{x,0,0.1}] using series theory. Any help?
Thanks!
$\displaystyle
\int_0^{0.1} e^{-x^2}dx
$
We will do this by expanding the integrand as a power series and then integrating term by term.
$\displaystyle
e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}
$
so:
$\displaystyle
e^{-x^2}=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}
$
Hence:
$\displaystyle
\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
$
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:
$\displaystyle
\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]
$
so:
$\displaystyle
\int_0^{0.1} e^{-x^2}dx= \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!(2n+1)} (0.1)^{2n+1} \right]
$
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