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Question
Compute the value of `int_0^(pi/2) sqrt(sinx+cosx) dx` usingTrapezoidal rule by dividing into six Subintervals.
Sum
Solution
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` |
Trapezoidal rule : `"I"=h/2[X+2R]` -----------------(1)
𝑿=𝒔𝒖𝒎 𝒐𝒇 𝒆𝒙𝒕𝒓𝒆𝒎𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔= 𝟐 𝑹=𝒔𝒖𝒎 𝒐𝒇 𝒓𝒆𝒎𝒂𝒊𝒏𝒊𝒏𝒈 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔=𝟓.𝟕𝟒𝟎𝟐
`"I"=pi/(12xx2)(2+2(5.7402))` ……………….(from 1)
I = 1.7636
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Numerical Integration‐ by Trapezoidal
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