Let us first deal with the function itself. The MacLaurin expansion has the shape $1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots$. Integrating from $0$ to $x$, we find that the Maclaurin expansion for the area up to $x$ has Maclaurin expansion $x-\frac{x^3}{3\cdot 1!}+\frac{x^5}{5\cdot 2!}-\frac{x^7}{7\cdot 3!}+\cdots$.
For the function, you were asked to truncate just after the $\frac{x^2}{1!}$ term. You can estimate the error in terms of the third derivative, evaluated at some unknown place $\xi$ between, in this case, $0$ and $0.5$. Now in fact this $\xi$ is close to $0$, but you do not officially know that, and the upper bound on the error that you get is quite a bit too pessimistic.
However, as you observed, the coefficient of $x^3$ is $0$, and therefore the Maclaurin polynomial up to $x^2$ is exactly the same as the Maclaurin polynomial up to $x^3$. So the error can be expressed in terms of the fourth derivative, and this gives a nicer upper bound on the error.
There is another way of looking at things that does not use the Lagrange formula for the remainder. Note that the Maclaurin polynomial, at least for $x\le 1$, is an alternating series. So the truncation error is less, in absolute value, than the first "neglected" term. This term has absolute value $\frac{x^4}{2!}$. So the absolute value of the error when we evaluate at $x=0.5$ is less than $\frac{(0.5)^2}{2!}$.
The error estimation for the integral uses exactly the same ideas.
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Best Answer
In general a Taylor series tells you what is happening near the point you're expanding around. To deal with far-away points, you'll want a different expansion, and in many cases that is available.
In many cases (including rational functions) there will be Laurent series that cover the whole real line except for some isolated points - you will study these if you take complex analysis.