Question

By trapezoidal rule, the approximate value of the integral $\underset{0}{\overset{6}{\int }}\frac{dx}{1+x^2}$ is

• A. $1.3128$
• B. $1.4108$
• C. $1.4218$
• D. None of these

Answer 1

Answer: (B)

Given that,
$I=\underset{0}{\overset{6}{\int }}\frac{dx}{1+x^2}$
From Eq. (i), $f(x)=\frac{1}{1+x^2}$
Now, divide the interval $[0,6]$ into six parts each of width
$h=\frac{6-0}{6}=1$
The value of $f(x)$ are given below

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$f(x)$	$1$	$0.5$	$0.2$	$0.1$	$0.0588$	$0.0385$	$0.027$

The trapezoidal rule is
$\underset{x_0}{\overset{x_0+nh}{\int }}ydx=\frac{h}{2}\left[\left(y_0+y_n\right)+2\left(y_1+y_2+y_3+\dots +y_{n-1}\right)\right]$
$\therefore \underset{0}{\overset{6}{\int }}\frac{1}{1+x^2}dx=\frac{1}{2}[(1+0.027)+2\{0.5+0.2+0.1+0.0588+0.0385]$
$=\frac{1}{2}[1.027+2(0.8973)]$
$=\frac{1}{2}[1.027+1.7946]$
$=\frac{1}{2}[2.8216]=1.4108$