Question

By simpson one third rule taking $n=4$, the value of the integral ${\int }_0^1\frac{1}{1+x^2}dx$ is equal to

• A. $.788$
• B. $0.781$
• C. $.785$
• D. None of these

Answer 1

The correct option is C $.785$

${\int }_0^1\frac{1}{1+x^2}dx$
According to simpson rule
${\int }_{x_0}^{x_n}ydx=\frac{h}{3}[y_0+4(y_1+y_3+y_5+....+y_{n-1})+(y_2+y_4+....+y_{n-2})+y_n]$
$\therefore x=1,h=\frac{x_n-x_0}{n}=\frac{1-0}{4}=\frac{1}{4}=.25$
Now $y=\frac{1}{1+x^2}$

	$x$	$y$
$1.$	$0$	$1$
$2.$	$.25$	$.941$	$\therefore y_0=1$
$3.$	$.50$	$.800$	$y_1=.941$
$4.$	$.75$	$.640$	$y_2=.80$
$5.$	$1.00$	$.500$	$y_3=.64$
			$y_4=0.5=y_n$

$\therefore {\int }_0^1\frac{1}{1+x^2}dx=\frac{.25}{3}[1+4(.941+0.64)+2(0.8)+.5]$
$=\frac{.25}{3}[1+1.581\times 4+1.600+.500]$
$=\frac{.25}{3}[1+6.324+2.100]$
$=\frac{.25}{3}\left[9.424\right]=\frac{9.424}{12}=.785$
Hence choice (c) is correct answer.