Chapter 6.5, Problem 18E

(a)

To determine

To calculate: The value of the integral using Simpson’s rule with n = 4 approximation

Answer to Problem 18E

The value of integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}=2.00456$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}$

Formula used:

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Calculation:

To approximate ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where [a, b] is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$

Now, apply the Simpson’s rule with

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

So,

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

  

Now, substitute the values from given table in this equation

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)}\\ =\frac{\frac{\pi }{4}}{3}\left(0+4\left(\frac{1}{\sqrt{2}}\right)+2\left(1\right)+4\left(\frac{1}{\sqrt{2}}\right)+0\right)\\ =\frac{1}{3}\left(0+4+2.828+6.928+2\right)\\ =\frac{\pi }{12}\left(0+2\sqrt{2}+2\sqrt{2}+0\right)\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}=2.00456\end{array}$

Conclusion:

The value of integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}=2.00456$

(b)

To determine

To calculate: The exact value of the integral is ${\displaystyle {\int }_0^{\pi }\sin xdx}$

Answer to Problem 18E

The value of the integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=2}$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}$

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

Calculation:

If f is continuous at every point of [a, b], and if F is any anti-derivative of f on [a, b], then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

This part of the fundamental theorem is also called the integral evaluation theorem.

Now, use the above formula to find the value of integral.

  $f\left(x\right)=\sin x$

If F is any anti-derivative of f, then

  $F\left(x\right)=-\cos x$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=\left(-\cos \pi -\left(-\cos 0\right)\right)}\\ =-\left(-1\right)+1\\ {\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=2}\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx=2}$