Q90 / 170 · Manipal (MET) 2020 · Mathematics
By trapezoidal rule, the approximate value of the integral $\int_0^6 \frac{d x}{1+x^2}$ is
- A$1.3128$
- Correct answerB$1.4108$
- C$1.4218$
- DNone of these
Solution
Given that,
$I=\int_0^6 \frac{d x}{1+x^2}$ ....(i)
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
$\begin{array}{c|c|c|c|c|c|c|c}\hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\hline f(x) & 1 & 0.5 & 0.2 & 0.1 & 0.0588 & 0.0385 & 0.027 \\\hline\end{array}$
The trapezoidal rule is
$\begin{aligned} \int_{x_0}^{x_0+n h} y d x= & \frac{h}{2}\left[\left(y_0+y_n\right)\right. \\ & \left.+2\left(y_1+y_2+y_3+\ldots+y_{n-1}\right)\right]\end{aligned}$
$\begin{aligned} \therefore \int_0^6 \frac{1}{1+x^2} d x & =\frac{1}{2}[(1+0.027)+2\{0.5+0.2 \\ & +0.1+0.0588+0.0385\}] \\ & =\frac{1}{2}[1.027+2(0.8973)] \\ & =\frac{1}{2}[1.027+1.7946] \\ & =\frac{1}{2}[2.8216]=1.4108\end{aligned}$
Hence, the correct answer is option: 'B'.