Q. By trapezoidal rule, the approximate value of the integral $\int\limits_{0}^{6} \frac{d x}{1+x^{2}}$ is
ManipalManipal 2020
Solution:
Given that,
$I=\int\limits_{0}^{6} \frac{d x}{1+x^{2}}$
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
$\int\limits_{x_{0}}^{x_{0}+n h} y d x=\frac{h}
{2}\left[\left(y_{0}+y_{n}\right)+2\left(y_{1}+y_{2}+y_{3}+\ldots+ y_{n-1}\right)\right]$
$\therefore \int\limits_{0}^{6} \frac{1}{1+x^{2}} d x=\frac{1}{2}[(1+0.027)+2\{0.5+0.2+0.1+0.0588+0.0385]$
$=\frac{1}{2}[1.027+2(0.8973)]$
$=\frac{1}{2}[1.027+1.7946]$
$=\frac{1}{2}[2.8216]=1.4108$
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| $f(x)$ | $1$ | $0.5$ | $0.2$ | $0.1$ | $0.0588$ | $0.0385$ | $0.027$ |