# Taylor series expansion approximating an integral?

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?

• Now, integrate each term of the series. Apr 14 '15 at 2:08

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 $0$).