Chapter 6.5, Problem 35E

To determine

To find:the trapezoidal approximation using 4 equal subdivisions.

Answer to Problem 35E

The value of the integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{\pi }{4}}\left(1+\sqrt{2}\right)$

Explanation of Solution

Given information:

The trapezoidal approximation of ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$ using 4 equal subdivisions of the interval of integration is

  $\begin{array}{l}A)\frac{\pi }{2}\\ B)\pi \\ C)\frac{\pi }{4}\left(1+\sqrt{2}\right)\\ D)\frac{\pi }{2}\left(1+\sqrt{2}\right)\\ E)\frac{\pi }{4}\left(1+\sqrt{2}\right)\end{array}$

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

Correct option is (C).

In the trapezoidal rule, if [ a, b ] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of f on [ 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,

  $\begin{array}{l}{y}_0=f\left(a\right)\\ y_1=f\left(x_1\right)\\ .\\ .\\ y_{n-1}=f\left(x_{n-1}\right)\\ y_n=f\left(b\right)\end{array}$

Now, apply the trapezoidal rule with

  $\begin{array}{l}n=4\\ f\left(x\right)=\sin x\\ \left[a,b\right]=\left[0,\pi \right]\end{array}$

So,

  $\begin{array}{l}h=\frac{b-a}{n}\\ =\frac{\pi -0}{4}\\ h=\frac{\pi }{4}\end{array}$

The table shows

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{h}{2}\left(y_0+2y_1+2y_2+2y_3+y_4\right)}\\ =\frac{\frac{\pi }{4}}{2}\left(0+2\left(\frac{1}{\sqrt{2}}\right)+2\left(1\right)+2\left(\frac{1}{\sqrt{2}}\right)+0\right)\\ =\frac{\pi }{8}\left(0+\sqrt{2}+2+\sqrt{2}+0\right)\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{\pi }{4}}\left(1+\sqrt{2}\right)\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{\pi }{4}}\left(1+\sqrt{2}\right)$