approximate ∫₃⁷ 1/ln(x) dx using the mid - point rule with three subdivisions to four decimal places. the…

Explanation:

Step1: Calculate the width of sub - intervals

The interval is $[a,b]=[3,7]$ and $n = 3$. The width of each sub - interval $\Delta x=\frac{b - a}{n}=\frac{7 - 3}{3}=\frac{4}{3}$.

Step2: Determine the sub - intervals and midpoints

The sub - intervals are $[3,3+\frac{4}{3}]=[3,\frac{13}{3}]$, $[\frac{13}{3},\frac{13}{3}+\frac{4}{3}]=[\frac{13}{3},\frac{17}{3}]$, $[\frac{17}{3},7]$. The midpoints are $x_1=\frac{3+\frac{13}{3}}{2}=\frac{\frac{9 + 13}{3}}{2}=\frac{11}{3}$, $x_2=\frac{\frac{13}{3}+\frac{17}{3}}{2}=5$, $x_3=\frac{\frac{17}{3}+7}{2}=\frac{\frac{17+21}{3}}{2}=\frac{19}{3}$.

Step3: Evaluate the function at the midpoints

Let $f(x)=\frac{1}{\ln(x)}$. Then $f(x_1)=\frac{1}{\ln(\frac{11}{3})}$, $f(x_2)=\frac{1}{\ln(5)}$, $f(x_3)=\frac{1}{\ln(\frac{19}{3})}$.

Step4: Apply the midpoint rule

The midpoint rule is $M_n=\Delta x\sum_{i = 1}^{n}f(x_i)$. So $M_3=\frac{4}{3}\left(\frac{1}{\ln(\frac{11}{3})}+\frac{1}{\ln(5)}+\frac{1}{\ln(\frac{19}{3})}\right)$. $\frac{1}{\ln(\frac{11}{3})}\approx\frac{1}{1.2993}\approx0.7697$, $\frac{1}{\ln(5)}\approx\frac{1}{1.6094}\approx0.6213$, $\frac{1}{\ln(\frac{19}{3})}\approx\frac{1}{1.8971}\approx0.5271$. $M_3=\frac{4}{3}(0.7697 + 0.6213+0.5271)=\frac{4}{3}(1.9181)\approx2.5575$.

Answer:

$2.5575$