| x |
y=x½+1 |
ydx (rectangle) |
| 0 |
1.00 |
- |
| ⅛ |
1.35 |
0.17 |
| ¼ |
1.50 |
0.19 |
| ⅜ |
1.61 |
0.20 |
| ½ |
1.71 |
0.21 |
| ⅝ |
1.79 |
0.22 |
| ¾ |
1.87 |
0.23 |
| ⅞ |
1.94 |
0.24 |
| 1 |
2.00 |
0.25 |
|
TOTAL AREA |
1.72 |
The table shows 8 equal intervals between 0 and 1, the y value for each of the x values and the area of the thin rectangle formed by combining the "dx" (thickness of the rectangle) as the width and y as the length. The sum of the areas comes to 1.72, compared with the true integral of 5/3=1.67 (2 places of decimals). This method uses the "outer" rectangles.
A better approximation can be found by calculating the area of the "inner" rectangles. The true integral lies in between the combined areas of the inner and outer rectangles, and the average of these areas should produce a better approximation. If this is done the inner area is 1.60, so the area lies between 1.60 and 1.72. The average is 1.66, very close to the true value.