How do you use a Power Series to estimate the integral #int_0^0.01e^(x^2)dx# ?

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Answer 1 · 2014-10-16T20:02:44

Since

#e^x=1+x+x^2/{2!}+cdots#,

#e^{x^2}=1+x^2+x^4/{2!}+cdots#

#int_0^{0.01}e^{x^2}dx=int_0^{0.01} (1+x^2+x^4/{2!}+cdots)dx#

#=[x+x^3/3+x^5/{10}+cdots]_0^{0.01}#

#=0.01+(0.01)^3/3+(0.01)^5/10+cdots#

#approx 0.01#

I hope that this was helpful.