The Function: $\int_0^{1/2}x^{2}\ln(1+x^2)dx$
So I tried approximating it doing the following. First, I figured out that the series of $\ln(1+x^2)$ is just $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{n+1}$. I've gotten it by taking the derivative of $\ln(1+x^2)$ and translated that into the geometric form of a series, and integrated that.
Now, I don't quite understand what allows me to simply take the first 4 terms and multiply them by $x^2$ to get the approximation I wanted (and obviously evaluating the bounds later). What allows us to completely ignore the fact that we haven't done anything with $x^2$ that we're simply allowed to use it on our end result now as is?
Would really appreciate some explanation, thanks in advance!
What the exercise asks you is to approximate the integrand function and then compute the integral of the approximation. Since an infinite series will be involved, you just take the first four terms of that series.
$$\int_0^{0.5} x^2\ln(1+x^2)\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k+1}\int_0^{0.5} x^{2n+4}\ \text{d}x = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k+1}\frac{2^{-2 k-5}}{2 k+5}$$
The first four terms of the series are thence
$$\frac{1}{160} -\frac{1}{1792} +\frac{1}{13824} -\frac{1}{90112} = \frac{3185513}{553512960} \approx 0.00575508(...)$$
Notice that
$$\int_0^{0.5} x^2\ln(1+x^2)\ \text{d}x = \frac{1}{72} \left(22+\ln \left(\frac{125}{64}\right)-48 \arctan\left(\frac{1}{2}\right)\right) = 0.0057548(...)$$
Final Remark
$x^2$ is a polynomial function, and for that reason it cannot be approximated with a power series, like Taylor series, in the same way you do for non polynomial function.