Chapter 6.5, Problem 34E

To determine

To calculate:The value of trapezoidal approximation of ${\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx$

Answer to Problem 34E

The value is ${\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx=\frac{1}{2}\left(e^4+2e^2+2e^0+e^{-2}\right)$

Explanation of Solution

Given information:

The trapezoidal approximation of ${\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx$

  $\begin{array}{l}A)e^4+e^2+e^0+e^{-2}\\ B)e^4+2e^2+2e^0+e^{-2}\\ C)\frac{1}{2}\left(e^4+2e^2+2e^0+e^{-2}\right)\\ E)\frac{1}{4}\left(e^4+2e^2+2e^0+e^{-2}\right)\end{array}$

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

Correct option is (D).

In the trapezoidal rule, if [ a, b ] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of fon [ a, b ] can be approximated by a straight line segment at each subinterval. The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases".

That is,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=h\left(\frac{y_0+y_1}{2}\right)+h\left(\frac{y_1+y_2}{2}\right)+\mathrm{....}+h\left(\frac{y_{n-1}+y_n}{2}\right)\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{....}+y_{n-1}+\frac{y_n}{2}\right)\\ =\frac{h}{2}\left(y_0+2y_1+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where

  $y_0=f\left(a\right),y_1=f\left(x_1\right)\mathrm{......}{y}_{n-1}=f\left(x_{n-1}\right),y_n=f\left(b\right)$

Now, apply the trapezoidal rule with

  $\begin{array}{l}n=3\\ f\left(x\right)=\frac{e^x}{2}\\ \left[a,b\right]=\left[-2,4\right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{4-\left(-2\right)}{3}\\ h=2\end{array}$

The following table gives the values of y for the corresponding values of x,

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx=\frac{h}{2}\left(y_0+2y_1+2y_2+y_3\right)\\ =\frac{2}{2}\left(\frac{e^{-2}}{2}+2\left(\frac{e^0}{2}\right)+2\left(\frac{e^2}{2}\right)+\left(\frac{e^4}{2}\right)\right)\\ {\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx=\frac{1}{2}\left(e^4+2e^2+2e^0+e^{-2}\right)\end{array}$

Conclusion:

The value is ${\displaystyle \underset{-2}{\overset{4}{\int }}\frac{e^x}{2}}dx=\frac{1}{2}\left(e^4+2e^2+2e^0+e^{-2}\right)$