Formula
1. Trapezoidal Rule
`int y dx = h/2 (y_0 + 2 (y_1 + y_2 + y_3 + ... + y_(n-1)) + y_n)`
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Examples
1. Find Solution using Trapezoidal rule
| x | f(x) |
| 1.4 | 4.0552 |
| 1.6 | 4.9530 |
| 1.8 | 6.0436 |
| 2.0 | 7.3891 |
| 2.2 | 9.0250 |
Solution:
The value of table for `x` and `y`
| x | 1.4 | 1.6 | 1.8 | 2 | 2.2 |
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| y | 4.0552 | 4.953 | 6.0436 | 7.3891 | 9.025 |
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Using Trapezoidal Rule
`int y dx = h/2 [y_0+y_4 + 2(y_1 + y_2 + y_3)]`
`int y dx = 0.2/2 [4.0552 + 9.025 + 2xx(4.953 + 6.0436 + 7.3891)]`
`int y dx = 0.2/2 [4.0552 + 9.025 + 2xx(18.3857)]`
`int y dx = 4.9852`
Solution by Trapezoidal Rule is `4.9852`
2. Find Solution 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 `y`
| x | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
|---|
| y | 1 | 0.9975 | 0.99 | 0.9776 | 0.8604 |
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Using Trapezoidal Rule
`int y dx = h/2 [y_0+y_4 + 2(y_1 + y_2 + y_3)]`
`int y dx = 0.1/2 [1 + 0.8604 + 2xx(0.9975 + 0.99 + 0.9776)]`
`int y dx = 0.1/2 [1 + 0.8604 + 2xx(2.9651)]`
`int y dx = 0.38953`
Solution by Trapezoidal Rule is `0.38953`
This material is intended as a summary. Use your textbook for detail explanation.
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