Estimate the error that results from using the Trapezoidal Rule with $n=4$ and with $n=6$ to approximate $\int_{1}^{2}\dfrac{e^{x}}{x}dx$. See Example 2.

Solution To estimate the error resulting from approximating $\int_{1}^{2}\dfrac{e^{x}}{x}dx$ using the Trapezoidal Rule, we need to find the absolute maximum value of $\vert f^{\prime \prime}\vert$ on the interval $[ 1,2] .$ We begin by finding $f^{\prime \prime}$: $$\begin{eqnarray*} f( x) &=&\dfrac{e^{x}}{x} \\[5pt] f^{\prime} (x) &=&\dfrac{d}{dx}\dfrac{e^{x}}{x}=\dfrac{xe^{x}-e^{x}}{x^{2}}= \dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}} \\[5pt] f^{\prime \prime} (x) &=&\dfrac{d}{dx}\left( \dfrac{e^{x}}{x}-\dfrac{e^{x}}{ x^{2}}\right) =\dfrac{d}{dx}\dfrac{e^{x}}{x}-\dfrac{d}{dx}\dfrac{e^{x}}{x^{2} }=\left( \dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}\right) -\dfrac{ e^{x}x^{2}-2e^{x}x}{x^{4}}\\[5pt] &=&\dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}-\dfrac{e^{x} }{x^{2}}+\dfrac{2e^{x}}{x^{3}} \\[5pt] &=&e^{x}\left( \dfrac{1}{x}-\dfrac{2}{x^{2}}+\dfrac{2}{x^{3}}\right) \end{eqnarray*}$$

Since $\vert f^{\prime \prime}\vert$ is continuous on the interval $[1,2] ,$ the Extreme Value Theorem guarantees that $\vert f^{\prime \prime}\vert$ has an absolute maximum on $[1,2]$. We find the absolute maximum of $\vert f^{\prime \prime}\vert$ using graphing technology. As seen from Figure 17, the absolute maximum is at the left endpoint $1.$ The absolute maximum value of $f^{\prime \prime}$ is $\left\vert e^{1}\left( \dfrac{1}{1}-\dfrac{ 2}{1^{2}}+\dfrac{2}{1^{3}}\right) \right\vert =e.$ So, $M=e$.

The absolute maximum of $\vert f^{\prime\prime} \vert$ on $[1,2 ]$ is at $x=1.$

When $n=4,$ the error in using the Trapezoidal Rule is $$\hbox{Error} \leq \dfrac{( b-a) ^{3}M}{12n^{2}}=\dfrac{(2-1)^{3}(e) }{(12)(4^{2})}=\dfrac{e}{192}\approx 0.014$$

That is, $$\begin{eqnarray*} 3.069-0.014 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.069+0.014 \\[5pt] 3.055 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.083 \end{eqnarray*}$$

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When $n=6,$ the error in using the Trapezoidal Rule is $$\hbox{Error} \leq \dfrac{\left( b-a\right) ^{3}M}{12n^{2}}=\dfrac{(2-1)^{3}(e) }{(12)(6^{2})}=\dfrac{e}{432}\approx 0.006$$

That is, $$\begin{eqnarray*} 3.063-0.006 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.063+0.006 \nonumber \\[5pt] 3.057 & \leq &\int_{1}^{2}\dfrac{e^{x}}{x}dx \leq 3.069 \end{eqnarray*}$$