Estimate error of $\int_0^{1/2} \exp(-x^3)\, dx$ Estimate within an error of at most $0.001$:
$$\int_0^{1/2} \exp(-x^3)\, dx$$
I am not sure if using Taylor Series is the best way to do this. Is there not a formula for estimating error of such an integral?
 A: Pencil and paper solution only:
Using the power series expansion of $\exp$:
$$I=\int_0^{1/2}\mathrm{e}^{-x^3}\,\mathrm{d}x=\sum_{n=0}^{+\infty}\int_0^{1/2}\frac{(-x)^{3n}}{n!}\,\mathrm{d}x=\sum_{n=0}^{+\infty}\frac{(-1)^n}{2^{3n+1}(3n+1)n!}.$$
Denote by $(S_N)_N$ and $(R_N)_N$ the sequences of partial sums and remainders of this series, namely:
$$\forall N\in\mathbb{N},\ S_N=\sum_{n=0}^N\frac{(-1)^n}{2^{3n+1}(3n+1)n!}\qquad\text{and}\qquad R_N=\sum_{n=N+1}^{+\infty}\frac{(-1)^n}{2^{3n+1}(3n+1)n!}.$$
Of course, for all $N\in\mathbb{N}$, $I=S_N+R_N$.
Now, the series we obtained for $I$ is an alternating series and since the sequence
$$\left(\frac1{2^{3n+1}(3n+1)n!}\right)_{n\in\mathbb{N}}$$
is decreasing, we have:
$$\forall N\in\mathbb{N},\ \lvert R_N\rvert\leq\frac1{2^{3N+4}(3N+4)(N+1)!}$$
(the absolute value of the remainder of the alternating series is non-greater than that of its first term). So it's enough to find $N\in\mathbb{N}$ such that
$$\frac1{2^{3N+4}(3N+4)(N+1)!}<0.001.$$
Taking $N=1$ works. In fact,
$$\lvert R_1\rvert\leq\frac1{2^77(2)!}=\frac1{2^87}<0.0006.$$
Also, using the fact that the sign of the remainder is that of its first term, which for $N=1$ is positive,
$$0\leq R_1<0.0006.$$
Now,
$$S_1=\frac12-\frac1{2^44}=\frac{31}{64},$$
Hence
$$\frac{31}{64}\leq I=S_1+R_1<\frac{31}{64}+0.0006.$$
By a straightforward long division, we obtain:
$$0.484<\frac{31}{64}<0.48438+0.0006=0.48498,$$
hence
$$0.484<I<0.48498.$$
(I purposely kept a few extra digits for the upper bound in order to have the value of $I$ correct to 3 decimal places: $I=0.484\ldots$).
