Question

4. Consider using the Simpson's 1/3 rule to estimate the following integral I[cos(x 3)l dx (a) Find the approximate values of 1 when the step size h-: 2 and h 1 , respectively. (b) Find an upper bound of the step size h in order to guarantee that the absolute error (in absolute value) of the estimate is less than 0.001. Hint: 2 sin x cos x = sin (2x). I cos x I " The arguments of all trigonometric functions are in radians." 1, and i sin x l s; 1 for all x.

Answer 1

(a)   $y=\cos ^2(x+3)$

let h=2, then

$x_o=0$

$y_o=\cos ^2(0+3)$

$y_o=0.98$

$x_1=2$

$y_1=\cos ^2(2+3)$

$y_1=0.08$

$x_2=4$

$y_2=\cos ^2(4+3)$

$y_2=0.568$

Now, according to Simpson's 1/3rd rule integration is given by

let h=1, then

$x_o=0$

$y_o=\cos ^2(0+3)$

$y_o=0.98$

$x_1=1$

$y_1=\cos ^2(1+3)$

$y_1=0.427$

$x_2=2$

$y_2=\cos ^2(2+3)$

$y_2=0.08$

$x_3=3$

$y_3=\cos ^2(3+3)$

$y_3=0.922$

$x_4=4$

$y_4=\cos ^2(4+3)$

$y_4=0.568$

Now, according to Simpson's 1/3rd rule integration is given by

$I=\frac{1}{3}[(0.98+0.568)+2(0.08)+4(0.427+0.922)]$

$I=2.368$

(b) error in Simpson's 1/3rd rule is given by

$E\leq \frac{(b-a)^5}{180n^4}[max\left | f^4(x) \right |]$

where   $max\left | f^4(x) \right |$ is the maximum value of the 4th order derivative of  

$f(x)=\cos ^2(x+3)$

$f^1(x)=-2\cos (x+3)\sin (x+3)$

$f^1(x)=-\sin (2x+6)$

$f^2(x)=-2\cos (2x+6)$

$f^3(x)=4\sin (2x+6)$

$f^4(x)=8\cos (2x+6)$

As,

$\left | \cos (2x+6) \right |\leq 1$ for all x.

So,  

$max\left | f^4(x) \right |=8$

It is given that

$a=0$ ,   $b=4$ and   $E\leq 0.001$   

So,

$E\leq \frac{(b-a)^5}{180n^4}[max\left | f^4(x) \right |]$

$\Rightarrow 0.001 =\frac{(4-0)^5}{180n^4}[8]$

$\Rightarrow n^4=\frac{1024}{180\times 0.001}[8]$

$\Rightarrow n^4=\frac{8192}{0.18}$

$\Rightarrow n^4=45511.11$

$\Rightarrow n=14.60$

Also,

$h=\frac{b-a}{n}$

$h=\frac{4-0}{14.60}$

$h=0.274$

So, h=0.274 is the maximum step size to get error less than 0.001.