Find the approximated integral value using Trapezoidal rule
| x | f(x) |
| 0.00 | 1.0000 |
| 0.25 | 0.9896 |
| 0.50 | 0.9589 |
| 0.75 | 0.9089 |
| 1.00 | 0.8415 |
Solution:The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=0` | `f(x_(0))=1` |
| `x_1=0.25` | `f(x_(1))=0.9896` |
| `x_2=0.5` | `f(x_(2))=0.9589` |
| `x_3=0.75` | `f(x_(3))=0.9089` |
| `x_4=1` | `f(x_(4))=0.8415` |
Method-1:Using Trapezoidal Rule
`int f(x) dx=(Delta x )/2 (f(x_(0))+2(f(x_(1))+f(x_(2))+f(x_(3))+...+f(x_(n-1)))+f(x_(n)))`
`int f(x) dx=(Delta x )/2 [f(x_(0))+2f(x_(1))+2f(x_(2))+2f(x_(3))+f(x_(4))]`
`f(x_(0))=1`
`2f(x_(1))=2*0.9896=1.9792`
`2f(x_(2))=2*0.9589=1.9178`
`2f(x_(3))=2*0.9089=1.8178`
`f(x_(4))=0.8415`
`int f(x) dx=0.25/2*(1+1.9792+1.9178+1.8178+0.8415)`
`=0.25/2*(7.5563)`
`=0.9445`
Solution by Trapezoidal Rule is `0.9445`
Method-2:Using Trapezoidal Rule
`int f(x) dx=(Delta x )/2 (f(x_(0))+2(f(x_(1))+f(x_(2))+f(x_(3))+...+f(x_(n-1)))+f(x_(n)))`
`int f(x) dx=(Delta x )/2 [f(x_(0))+f(x_(4))+2(f(x_(1))+f(x_(2))+f(x_(3)))]`
`=0.25/2 [1 +0.8415 + 2xx(0.9896+0.9589+0.9089)]`
`=0.25/2 [1 +0.8415 + 2xx(2.8574)]`
`=0.25/2 [1 +0.8415 + 5.7148]`
`=0.9445`
Solution by Trapezoidal Rule is `0.9445`
This material is intended as a summary. Use your textbook for detail explanation.
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