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Explanation:

Step1: Calculate the width of sub - intervals

The formula for the width $\Delta x$ of sub - intervals is $\Delta x=\frac{b - a}{n}$, where $a = 1$, $b = 5$, and $n=6$. So, $\Delta x=\frac{5 - 1}{6}=\frac{4}{6}=\frac{2}{3}$.

Step2: Determine the endpoints of sub - intervals

The endpoints $x_i=a + i\Delta x$, for $i = 0,1,\cdots,6$. So $x_0 = 1$, $x_1=1+\frac{2}{3}=\frac{5}{3}$, $x_2=1 + 2\times\frac{2}{3}=\frac{7}{3}$, $x_3=1+3\times\frac{2}{3}=3$, $x_4=1 + 4\times\frac{2}{3}=\frac{11}{3}$, $x_5=1+5\times\frac{2}{3}=\frac{13}{3}$, $x_6=5$.

Step3: Evaluate the function at endpoints

Let $f(x)=\frac{\cos(x)}{x}$. Then $f(x_0)=f(1)=\cos(1)$, $f(x_1)=\frac{\cos(\frac{5}{3})}{\frac{5}{3}}=\frac{3\cos(\frac{5}{3})}{5}$, $f(x_2)=\frac{\cos(\frac{7}{3})}{\frac{7}{3}}=\frac{3\cos(\frac{7}{3})}{7}$, $f(x_3)=\frac{\cos(3)}{3}$, $f(x_4)=\frac{\cos(\frac{11}{3})}{\frac{11}{3}}=\frac{3\cos(\frac{11}{3})}{11}$, $f(x_5)=\frac{\cos(\frac{13}{3})}{\frac{13}{3}}=\frac{3\cos(\frac{13}{3})}{13}$, $f(x_6)=f(5)=\frac{\cos(5)}{5}$.

Step4: Apply Simpson's rule

Simpson's rule is $\int_{a}^{b}f(x)dx\approx\frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)]$. Substitute $\Delta x=\frac{2}{3}$ and the function values: [ \begin{align*} &\frac{\frac{2}{3}}{3}[\cos(1)+4\times\frac{3\cos(\frac{5}{3})}{5}+2\times\frac{3\cos(\frac{7}{3})}{7}+4\times\frac{\cos(3)}{3}+2\times\frac{3\cos(\frac{11}{3})}{11}+4\times\frac{3\cos(\frac{13}{3})}{13}+\frac{\cos(5)}{5}]\ =&\frac{2}{9}[\cos(1)+\frac{12\cos(\frac{5}{3})}{5}+\frac{6\cos(\frac{7}{3})}{7}+\frac{4\cos(3)}{3}+\frac{6\cos(\frac{11}{3})}{11}+\frac{12\cos(\frac{13}{3})}{13}+\frac{\cos(5)}{5}] \end{align*} ] Using a calculator: $\cos(1)\approx0.54$, $\cos(\frac{5}{3})\approx - 0.08$, $\cos(\frac{7}{3})\approx - 0.90$, $\cos(3)\approx - 0.99$, $\cos(\frac{11}{3})\approx0.72$, $\cos(\frac{13}{3})\approx - 0.42$, $\cos(5)\approx0.28$. [ \begin{align*} &\frac{2}{9}[0.54+\frac{12\times(- 0.08)}{5}+\frac{6\times(-0.90)}{7}+\frac{4\times(-0.99)}{3}+\frac{6\times0.72}{11}+\frac{12\times(-0.42)}{13}+\frac{0.28}{5}]\ =&\frac{2}{9}[0.54 - 0.192-0.771 - 1.32+0.393 - 0.397+0.056]\ =&\frac{2}{9}[0.54+0.393 + 0.056-(0.192 + 0.771+1.32 + 0.397)]\ =&\frac{2}{9}[0.989 - 2.68]\ =&\frac{2}{9}\times(-1.691)\ \approx - 0.38 \end{align*} ]

Answer:

$-0.38$