Evaluate \[ I=\int_0^{1/2}\frac{\sin x}{x}\,dx \] to an accuracy of two decimal places: find \(I^*\) such that \[ |I-I^*|<0.005. \]
Solution
1. For every \(x\in[0,1/2]\),\[ \sin x = x-\frac{x^3}{6}+R(x), \qquad \abs{R(x)} \leq \frac{x^4}{24}. \]
Taylor’s theorem about \(0\), through degree \(3\), gives \[ \sin x = x-\frac{x^3}{6} + \frac{\sin \xi_x}{24}x^4 \] for some \(\xi_x\) between \(0\) and \(x\). Since \[ \abs{\sin \xi_x}\leq1, \] the stated remainder bound follows.
2. If\[ I_0 \coloneqq \int_0^{1/2} \left(1-\frac{x^2}{6}\right)\,dx, \]then\[ I_0 = \frac12-\frac1{144} = \frac{71}{144}. \]
Direct integration gives \[ \begin{aligned} I_0 &= \left[ x-\frac{x^3}{18} \right]_{0}^{1/2}\\ &= \frac12-\frac{1}{8\cdot18}\\ &= \frac12-\frac1{144}. \end{aligned} \]
3. The approximation error satisfies\[ \abs{I-I_0} \leq \frac1{1536}. \]
For \(x>0\), step 1 gives \[ \frac{\sin x}{x} = 1-\frac{x^2}{6} + \frac{R(x)}x, \] with \[ \abs{\frac{R(x)}x} \leq \frac{x^3}{24}. \] The same bound extends continuously to \(x=0\). Therefore \[ \begin{aligned} \abs{I-I_0} &\leq \int_0^{1/2}\frac{x^3}{24}\,dx\\ &= \left[ \frac{x^4}{96} \right]_{0}^{1/2}\\ &= \frac{1}{16\cdot96}\\ &= \frac1{1536}. \end{aligned} \]
4. The number\[ \boxed{I^*=0.49} \]satisfies\[ \abs{I-I^*}<0.005. \]
By steps 2 and 3, \[ \begin{aligned} \abs{I-0.49} &\leq \abs{I-I_0} + \abs{I_0-0.49}\\ &\leq \frac1{1536} + \left( \frac{71}{144}-\frac{49}{100} \right)\\ &= \frac1{1536} + \frac{11}{3600}. \end{aligned} \] Moreover, \[ \frac1{1536}<\frac1{1500} \qquad\text{and}\qquad \frac{11}{3600}<\frac{11}{3300}=\frac1{300}, \] so \[ \abs{I-0.49} < \frac1{1500}+\frac1{300} = \frac1{250} = 0.004 < 0.005. \]
5. Q.E.D.
Step 4 gives the required two-decimal approximation with the stated strict error tolerance.