Question

Compute the value of `int_0^(pi/2) sqrt(sinx+cosx) dx` using Simpson’s (3/8)^th rule by dividing into six Subintervals.

Answer 1

Let I = `int_0^(pi/2) sqrt(sinx+cosx) dx` 

Dividing limits into 6 subintervals . n=6

a = 0, `b=pi/2    thereforeh=(b-a)/n=pi/12`

`x_0=0`	`x_1=pi/12`	`x_2=(2pi)/12`	`x_3=(3pi)/12`	`x_4=(4pi)/12`	`x_5=(5pi)/12`	`x_6=(6pi)/12`
`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`

Simpson’s (𝟑/𝟖)^𝒕𝒉 rule :

`"I"=(3h)/8[X+2T+3R]`  -------------(1)

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 "I"=(3xxpi)/(8xx12)[2+2xx1.1892+3xx4.551]`

∴ I = 1.7702