Answer
$17.402367$
Work Step by Step
$\int_{1}^{4} \frac{e^x}{x}\, dx$
Using Simpson's Rule with $n=6$ sub-intervals.
$ a=1 , b=4, n=6 $
$h=\frac{b-a}{n} =\frac{4-1}{6} = 0.5$
Nodes: $x_{0}=1, x_{1}=1.5, x_{2}=2, x_{3}=2.5,x_{4}=3,x_{5}=3.5,x_{6}=4$
Since: $f(x)=\frac{e^x}{x}$
Apply Simpson's Rule:
$S_{n}= \frac{h}{3}[f(x_{0})+4\sum_{\text {odd,} i} f(x_{i})+ 2 \sum_{\text {even,} i\ne0,n}f(x_{i}) + f(x_{n}) $
Odd indices : $i = 1,3,5 $
$4(f(x_{1})+f(x_{3})+f(x_{5}))$
$=4(\frac{e^{1.5}}{1.5}+\frac{e^{2.5}}{2.5}+\frac{x^{3.5}}{3.5})$
$=4(16.816721)$
$=67.266884$
Even Indices (where $i\ne0,n$) : $i=2,4$
$2(f(x_{2}) + f(x_{4}))$
$=2(\frac{e^2}{2} + \frac{e^3}{3}) $
$= 2(10.389707)$
$=20.779414$
End Points: $i=0,6$
$f(x_{0})+f(x_{6})= \frac{e^1}{1}+\frac{e^4}{4} = 16.367904$
Combined:
$S_{6} \approx \frac{0.5}{3}(16.367904+ 67.266884+20.779414)$
$S_{6}\approx17.402367$
