Question

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

• A. $0.788$
• B. $0.781$
• C. $0.785$
• D. None of these

Answer 1

Answer: (C)

$0.785$

Here, $h=\frac{1}{4},0.25,y=\frac{1}{1+x^2}$

	x	y
1	$0$	$1.0$
2	$0.25$	$0.941$
3	$0.25$	$0.941$
4	$0.75$	$0.64$
5	$1$	$0.5$

By Simpson's Rule:
${\int }_0^1\frac{dx}{1+x^2}=\frac{1}{4\times 3}$
$\left[\right(1+0.5)+4(0.941+0.941+0.64)+2(0.8\left)\right]$
$=\frac{1}{12}\left[9.424\right]=0.785$