<s>[INST] Evaluate $ \int_0^1 \sqrt{x}\sin(x)dx $ to accuracy 0.001. Evaluate
$$
\int_0^1 \sqrt{x}\sin(x)dx
$$
to accuracy 0.001.
By definition, there exists an N such that for n > N,
$$
\left| \int_0^1 \sqrt{x}\sin(x)dx - \sum\limits_{i=1}^n \sqrt{x^{*}_i}\sin(x^{*}_i) \Delta x\right| <0.001.
$$
But from there I do not know how to proceed.  Are there any theories regaurding approximation that would be helpful?
 [/INST] Using @Mohammad Ali Baydoun: ideea, the integral lies between the increasing sequence
$$ \int_0^1  \sqrt{x}\cdot (x - \frac{x^3}{6}) dx, \int_0^1  \sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} - \frac{x^7}{7!})\,dx, \ldots $$
and the decreasing sequence 
$$ \int_0^1  \sqrt{x}\cdot x\,  dx, \int_0^1  \sqrt{x}\cdot ( x- \frac{x^3}{6}+ \frac{x^5}{120})\,dx, \ldots $$
These are sequences of rational numbers, easy to evaluate,  that converge to the same limit $\int_0^1 \sqrt{x} \sin x dx$.
Note that the inequalities between $\sin x$ and its odd and even order Taylor approximations are valid for all positive $x$, something that does not follow right away from Leibniz estimate for alternating series with decreasing absolute values of terms. See the picture below: 

The fifth and the seventh order Taylor approximation already give the correct fourth  decimal :
$$\frac{608527}{1670760} < \int_0^1 \sqrt{x} \sin x dx< \frac{2557}{7020}$$
The approximation on both sides
$$\sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} - \frac{x^7}{7!})\le \sqrt{x} \sin x \le \sqrt{x}\cdot (x - \frac{x^3}{6}+ \frac{x^5}{120} )$$ on $[0,1]$ has an error less than $\frac{1}{7!} = \frac{1}{5040}< \frac{1}{5000} = 0.0002$ so the estimates for the integrals are apriori better than $0.0002$ since the interval has length $1$. It turns out that the estimates are better than $0.0001$, as we see from evaluating both of the fractions. 
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