This is my question with context (I posted earlier but needed to change it!)
Calculate: $$\int_0^\frac{1}{2} e^{-x^2} \,dx$$
by using a Taylor approximation of $e^{−x^2}$ up to the term in $x^4$. Then I need to estimate the error, using the correct remainder term, and tell whether the answer will be correct to 2 decimal places.
The first part I did fine:
$\int_0^\frac{1}{2} T_4(x) \,dx=\int_0^\frac{1}{2} 1-x^2+\frac{x^4}{2!} \,dx$
Which gives $\frac{443}{960}$. But then I am not sure how to go about estimating the error? I have the formula:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1}$$
If anyone can help show me how to do the remainder part, this would be much appreciated!
Best Answer
The next term in the expansion of the integrand is $-\frac{x^6}{3!}$ Since it is an alternating series, the remainder is $\le \int_0^\frac{1}{2} max\frac{x^6}{3!}dx=\frac{2^{-7}}{3!}=0.00130208333$.