Find the approximated integral value using Trapezoidal rule
| x | f(x) |
| 0.0 | 1.0000 |
| 0.1 | 0.9975 |
| 0.2 | 0.9900 |
| 0.3 | 0.9776 |
| 0.4 | 0.8604 |
Solution:The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=0` | `f(x_(0))=1` |
| `x_1=0.1` | `f(x_(1))=0.9975` |
| `x_2=0.2` | `f(x_(2))=0.99` |
| `x_3=0.3` | `f(x_(3))=0.9776` |
| `x_4=0.4` | `f(x_(4))=0.8604` |
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*0.9975=1.995`
`2f(x_(2))=2*0.99=1.98`
`2f(x_(3))=2*0.9776=1.9552`
`f(x_(4))=0.8604`
`int f(x) dx=0.1/2*(1+1.995+1.98+1.9552+0.8604)`
`=0.1/2*(7.7906)`
`=0.3895`
Solution by Trapezoidal Rule is `0.3895`
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)))]`
`=0.1/2 [1 +0.8604 + 2xx(0.9975+0.99+0.9776)]`
`=0.1/2 [1 +0.8604 + 2xx(2.9651)]`
`=0.1/2 [1 +0.8604 + 5.9302]`
`=0.3895`
Solution by Trapezoidal Rule is `0.3895`
This material is intended as a summary. Use your textbook for detail explanation.
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