Question

The table below gives values of a function f(x) obtained for values of x at intervals of 0.25.

X	0	0.25	0.5	0.75	1.0
f(X)	1	0.9412	0.8	0.64	0.50

The value of the integral of the fucntion between the limits 0 to 1 using Simpson's rule is

• A. 0.7854
• B. 2.3562
• C. 3.1416
• D. 7.5000

Answer 1

The correct option is A 0.7854

Given,

X	0	0.25	0.5	0.75	1.0
f(X)	1	0.9412	0.8	0.64	0.50
	$y_0$	$y_1$	$y_2$	$y_3$	$y_4$

simpson's $\left(\frac{1}{3}rd\right)$ rule
${\int }_0^{1.0}f\left(x\right)dx=\frac{h}{3}\left[\right(y_0+y_4+4(y_1+y_3)+2\left(y_2\right)]$
$=\frac{0.25}{3}\left[\right(1+0.5)+4(0.9412+0.64)+2(0.8\left)\right]$
= 0.7854