Question

By Simpson rule taking $n=4$, the value of the integral $\int_{0}^{1} \frac{1}{1+x^{2}} d x$ is equal to

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

Answer 1

Answer: (C)

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

	x	y
1	0	1.0
2	0.25	0.941
3	0.5	0.8
4	0.75	0.64
5	1	0.5

By Simpson's Rule
$\int_\limits{0}^{1} \frac{d x}{1+x^{2}}=\frac{1}{4 \times 3}$
$[(1+0.5) + 4(0.941+ 0.64)+2(0.8)]$
$=\frac{1}{12}[9,424]=0.785$