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I need to use the Taylor series expansion of $$\frac{1}{1+3x^2} $$ to find a series approximating $$\int_0^1 \frac{1}{1+3x^2} \, dx $$ and $$\int_0^{1/3} \frac{1}{1+3x^2} \, dx $$ I tried to start the problem using the Maclaurin series for $$\frac{1}{1+x}$$ and plugged in 3x^2 for x.

Edit: After taking the integral, how do I show that the approximation will work for the second integral, but will not for the first integral?

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  • $\begingroup$ Now, integrate each term of the series. $\endgroup$ – Tim Raczkowski Apr 14 '15 at 2:08
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The big key here is this: say you have a function $f(x)$, and that it has a power series representation $$ f(x)=\sum_{n=0}^{\infty}a_nx^n, \qquad\lvert x\rvert < R $$ (where $R$ is the radius of convergence). Then it turns out that any antiderivative $F(x)$ of $f(x)$ can be written $$ F(x)=C+\sum_{n=0}^{\infty}a_n\frac{x^{n+1}}{n+1},\qquad \lvert x\rvert<R, $$ for some choice of the constant $C$. (Notably: this is the term-by-term integral of the original power series.)

In particular, this means that if $a,b\in(-R, R)$, then $$ \int_a^b f(x)\,dx=F(b)-F(a)=\sum_{n=0}^{\infty}a_n\frac{b^{n+1}}{n+1}-\sum_{n=0}^{\infty}a_n\frac{a^{n+1}}{n+1}. $$ In this particular case, we have $a=0$ and $b=0$, so that (assuming the radius of convergence exceeds $1$), we have $$ \int_0^1 f(x)\,dx=\sum_{n=0}^{\infty}\frac{a_n}{n+1} $$ (where the second term disappears because each has a positive power of $0$).

So, if you can find a power series representation for your function (which you had a sound approach for finding above), you can follow this to come up with a power series representation for the integral. But, if you can find a power series representing the integral, you can approximate its value to approximate the desired integral as well.

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