Use the Trapezoidal Rule with $n=4$ and $n=6$ to approximate $\int_{1}^{2} \dfrac{e^{x}}{x}dx$. Express the answer rounded to three decimal places.

Solution  We begin by partitioning the interval $[1,2]$ into four subintervals, each of width $\Delta x=\dfrac{2-1}{4}=\dfrac{1}{4}$: $$\left[ 1,\,\dfrac{5}{4}\right]\qquad \left[ \dfrac{5}{4},\,\dfrac{3}{2} \right]\qquad \left[ \dfrac{3}{2},\,\dfrac{7}{4}\right]\qquad \left[ \dfrac{7}{4},\,2\right]$$

The values of $f ( x) =\dfrac{e^{x}}{x}$ corresponding to each endpoint are $$\begin{eqnarray*} f(1) &=& e \quad f\!\left( \dfrac{5}{4}\right) = \dfrac{4e^{5/4}}{5}\quad f\!\left( \dfrac{3}{2}\right) =\dfrac{2e^{3/2}}{3}\quad f\!\left( \dfrac{7}{4}\right) = \dfrac{4e^{7/4}}{7}\quad f( 2) =\dfrac{e^{2}}{2} \end{eqnarray*}$$

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Then, using the Trapezoidal Rule, we get $$\int_{1}^{2}\dfrac{e^{x}}{x}dx\approx \dfrac{1}{2\cdot 4}\,\left[ e+2\cdot \dfrac{4e^{5/4}}{5}+2\,\cdot \dfrac{2e^{3/2}}{3}+2\,\cdot \dfrac{4e^{7/4}}{7} +\dfrac{e^{2}}{2}\right] \approx 3.069$$

To approximate $\int_{1}^{2}\dfrac{e^{x}}{x}\,dx$ using six subintervals, we partition $[1,2]$ into six subintervals, each of width $\Delta x= \dfrac{2-1}{6}=\dfrac{1}{6}$: $$\left[ 1,\,\dfrac{7}{6}\right]\! \quad \left[ \dfrac{7}{6},\,\dfrac{4}{3} \right]\! \quad \left[ \dfrac{4}{3},\,\dfrac{3}{2}\right]\! \quad \left[ \dfrac{3}{2},\dfrac{5}{3}\right]\!\quad \left[ \dfrac{5}{3},\dfrac{11}{6} \right]\!\quad \left[ \dfrac{11}{6},2\right]$$

Then, using the Trapezoidal Rule, we get $$\begin{eqnarray*} \int_{1}^{2}\dfrac{e^{x}}{x}dx\approx \dfrac{1}{2\cdot 6}\,\left[f(1)+2f\! \left( \dfrac{7}{6}\right) +2f\! \left( \dfrac{4}{3}\right) \right.&+& 2f\! \left( \dfrac{3}{2}\right) +2f\! \left( \dfrac{5}{3}\right) \\[5pt] &+& \left. 2f\!\left( \dfrac{11}{6}\right) +f(2) \right] \approx 3.063 \end{eqnarray*}$$