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Question
Compute the value of `int _0.2^1.4 (sin x - In x+e^x) ` 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 _0.2 ^1.4 (sin x -"in" x+e^x) dx`
∴` n=6 ∴ h=(b-a)/n=(1.4-0.2)/6=1/5`
| `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` |
| `y_0=3.02` | `y_1=2.79` | `y_2=2.89` | `y_3=3.16` | `y_4=3.55` | `y_5=4.06` | `y_6=4.4` |
(i) Trapezoidal rule : `I= h/2 [x+2R]` -----------------(1)
X = sum of extreme ordinates=`7.42`
R=sum of remaining ordinates = `16.45`
`I=1/5xx2 (7.42+2(16.45))` ……………….(from 1)
`I=4.032`
(ii) Simpson’s `(1/3)^(rd)` rule :
`I=h/3[X+2E+40]` ---------------(2)
X= sum of exterme ordinates= `y_0+y_6=4.4+3.02=7.42`
E= 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=1/3xx5(7.42+2xx6.44+4xx10..01)`
`I = 4.022 `
(iii) Simpson’s `(3/8)^(th)` rule
`I=3h/8[X+2T+3R]` -------------(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`
∴ `I=(3xx1)/(8xx5) [7.42+2xx3.16+3xx13.49]`
`[I=4.02075]`
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