$\int^{\pi/2}_0 \sqrt{sinx + cosx} \ dx$ using (i) Trapezoidal rule (ii) Simpson’s $(1/3)^{rd}$ rule (iii) Simpson’s $(3/8)^{th}$ rule by dividing into six subintervals

Solution:

Let $I=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x+\cos x} d x$

dividing limits into 6 subintervals. $n=6$

$a=0, b=\frac{\pi}{2} \quad \therefore h=\frac{b-a}{n}=\frac{\pi}{12}$

$\begin{array}{|c|c|c|c|c|c|}\hline x_{0} {=0} & {x_{1}=\pi / 12} & {x_{2}=2 \pi / 12} & {x_{3}=3 \pi / 12} & {x_{4}=4 \pi / 12} & {x_{5}} & {x_{6}} \\ \hline y_{0=1} & {y_{1}=1.1067} & {y_{2}=1.1688} & {y_{3=1.1892}} & {y_{4=1.1688}} & {y_{5=1.1067}} & {y_{6}=1} \\ \hline\end{array}$

(i) Trapezoidal rule:

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

$\quad X=$ sum of extreme ordinates $=2$

$R=$ sum of remaining ordinates $=5.7402$

$I=\frac{\pi}{12 \times 2}(2+2(5.7402))----from(1)$

$\quad I=1.7636$

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

$\mathrm{I}=\frac{h}{3}[X+2 E+40]----(2)$

$X=$ sum of extreme ordinates $=y_{0}+y_{6}=1+1=2$

$E=$sum of even base ordinates $=y_{2}+y_{4}=2.3376$

$o=$sum of odd base ordinates $=y_{1}+y_{3}+y_{4}=3.4026$

$1=\frac{\pi}{3 \times 12}(2+2 \times 2.3376+4 \times 3.4026)----from(2)$

$I=1.7693$

(iii) Simpson's $(3 / 8)$ th rule:

$I=\frac{3 h}{8}[X+2 T+3 R]----(3)$ $X=$sum of extreme ordinates$=y_{0}+y_{6}=0+0.5=2$

$T=$sum of multiple of three base ordinates $=y_{3}=1.1892$

$R=$sum of remaining ordinates$=y_{1}+y_{2}+y_{4}+y_{5}=4.551$

$\therefore \mathrm{I}=\frac{3 \times \pi}{8 \times 12}[2+2 \times 1.1892+3 \times 4.551]$

$\therefore \mathrm{I}=1.7702$