Q114 / 170 · Manipal (MET) 2020 · Mathematics
By Simpson's $\frac{1}{3}$ rd rule, the approximate value of the integral $\int_1^2 e^{-x / 2} d x$ using four intervals, is
- A$0.377$
- B$0.487$
- Correct answerC$0.477$
- D$0.387$
Solution
Given integral is $\int_1^2 e^{-x / 2} d x$
On dividing interval $[1,2]$ in four parts, we have
$h=\frac{2-1}{4}=\frac{1}{4}$
Now, value of $f(x)=e^{-x / 2}$ is given below
$\begin{array}{c|c|c|c|c|c}\hline x & 1 & \frac{5}{4} & \frac{3}{2} & \frac{7}{4} & e \\\hline f(x) & 0.6065 & 0.5352 & 0.4724 & 0.4168 & 0.3679 \\\hline\end{array}$
Now, Simpson's $\frac{1}{3}$ rule is
$\begin{aligned} \int_{x_0}^{x_0+n h} f(x) d x= & \frac{h}{3} \\ & {\left[\left(y_0+y_n\right)+4\left(y_1+y_3+\ldots+y_{n-1}\right)\right.} \\ & \left.+2\left(y_2+y_4+\ldots+y_{n-2}\right)\right]\end{aligned}$
$\begin{aligned} \therefore \int_1^2 e^{-x / 2} d x= & \frac{1}{12}[(0.6065+0.3679) \\ & +4(0.5352+0.4168)+2(0.4724)]\end{aligned}$
$\begin{aligned} & =\frac{1}{12}[0.9744+3.808+0.9448] \\ & =\frac{1}{12} \times 5.7272=0.477\end{aligned}$
Hence, the correct answer is option: 'C'.