Evaluate $ \int_0^1 \sqrt{x}\sin(x)dx $ to accuracy 0.001.

Question

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?

Answer 1

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.

Answer 2

Hint

Use the Taylor series expansion of $\sin x$ and do term wise integration. $$\sqrt{x}\sin x = \sum_{k=0}^{\infty} (-1)^k\frac{x^{2k+\frac{3}{2}}}{(2k+1)!}.$$ To get the desired accuracy you will need to adjust the number of terms you want to use.

Answer 3

$\text{Hints: }$

$$\sqrt x \sin x = \sqrt x\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x^{1.5} - \frac{x^{3.5}}{3!} + \frac{x^{5.5}}{5!} - \frac{x^{7.5}}{7!} + \dots$$

$$| \text{error} | \le |\text{First unused term}|$$

$\text{Your integral is then simple to evaluate using }$

$$\int_a^b x^kdx = \left[ \frac{x^{k+1}}{k+1} \right]_a^b$$