Using alternating series estimation theorem with non-zero lower integration limits

[DivGradCurl]

Consider the following:

$$\int _0 ^1 \sqrt{1+x^4} \mbox{ } dx = \left[ x + \frac{x^5}{2\cdot 5} - \frac{1}{2!2^2 9}x^8 + \frac{1\cdot 3}{3!2^3 13}x^{12} - \frac{1\cdot 3\cdot 5}{4!2^4 17}x^{16} +\dotsb \right] _0 ^1$$

According to the alternating series estimation theorem, we find:

$$\left| R_n \right| \leq b_{n+1} < \left| \mbox{ error } \right| \Longrightarrow \frac{1\cdot 3}{3!2^3 13} < 10^{-2} \Longrightarrow \int _0 ^1 \sqrt{1+x^4} \mbox{ } dx \approx 1 + \frac{1}{2\cdot 5} - \frac{1}{2!2^2 9} \approx 1.09$$

The limits up there are easy to work with. So, how about if we'd had

$$\int _{0.7} ^{1.5} \sqrt{1+x^4} \mbox{ } dx$$

instead? Do we need to take into account those limits when applying the alternating series estimation theorem? I mean:

$$\left| R_n (x) \right| \leq b_{n+1} (x) < \left| \mbox{ error } \right|$$

Thanks​

 

[vela]

It seems pretty clear you'd have to take into account the limits, doesn't it? Different limits result in different alternating series so different error estimates from the nth term.

In your hypothetical case, there is another complication in that if you just plugged in the new limits to the series about $x=0$, the series wouldn't converge for the upper limit. However, if you were to expand the integrand about $x=0.7$, the series would converge over the entire range of integration, and you could just plug in the limits after integration. Clearly, in this case, dealing with the limits results in a different series from your original example.