$\int^{1.4}_{0.2} (sin x – ln x + e^x) dx$ using (i) Trapezoidal rule (ii) Simpsons $(1/3)^{rd}$ rule (iii) Simpsons $(3/8)^{th}$ rule by dividing into six subintervals.

Solution:

let $I=\int_{0.2}^{1.4}\left(\sin x-\ln x+e^{x}\right) d x$

Dividing limits in six subintervals.

$\quad \therefore \mathrm{n}=6 \quad \therefore \mathrm{h}=\frac{b-a}{n}=\frac{1.4-0.2}{6}=\frac{1}{5}$

$\begin{array}{|c|c|c|c|c|c|c|}\hline x_{0=0.2} & {x_{1}=0.4} & {x_{2}=0.6} & {x_{3}=0.8} & {x_{4}=1.0} & {x_{5}=1.2} & {x_{6}=1.4} \\ \hline y_{0=3.02} & {y_{1}=2.79} & {y_{2}=2.89} & {y_{3}=3.16} & {y_{5}=3.55} & {y_{5}=4.06} & {y_{6}=4.4} \\ \hline\end{array}$

(i) Trapezoidal rule:

$\mathrm{I}=\frac{h}{2}[X+2 R]----(1)$

$X=$ sum of extreme ordinates $=7.42$

$R=$ sum of remaining ordinates $=16.45$

$I=\frac{1}{5 \times 2}(7.42+2(16.45))---from(1)$

$I=4.032$

(ii) $\quad$ Simpson's $(1 / 3)^{r d}$ rule:

$I=\frac{h}{3}[X+2 E+4 O]---(2)$

$X=$sum of extreme ordinates$=y_{0}+y_{6}=4.4+3.02=7.42$

$E=\operatorname{sum~of~even~base~ordinates}=y_{2}+y_{4}=6.44$

$O=$sum of odd base ordinates $=y_{1}+y_{3}+y_{5}=10.01$

$I=\frac{1}{3 \times 5}(7.42+2 \times 6.44+4 \times 10.01)----------from(2)$

$I=4.022$

(iii) Simpson's $(3 / 8)^{t h}$ rule:

$\mathrm{I}=\frac{3 h}{8}[X+2 T+3 R]----(3)$

$X=$ sum of extreme ordinates $=y_{0}+y_{6}=4.4+3.02=7.42$

$T=$sum of multiple of three base ordinates $=y_{3}=3.16$

$R=$ sum of remaining ordinates $=y_{1}+y_{2}+y_{4}+y_{5}=13.49$

$\therefore I=\frac{3 \times 1}{8 \times 5}[7.42+2 \times 3.16+3 \times 13.49]$

$\therefore I=4.02075$