Area Approximation without Graph

Jason76

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Post Edited Two Times: 9/3/2013

No question here. But if somebody sees something wrong, will post another question pertaining to specific rule:

Here is an easier way to do Simpson's rule without looking at a graph first:

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + 3i\Delta x) - 3\Delta x )]\).

\(\displaystyle \Delta x = \dfrac{b - a}{3n}\)

Interval: \(\displaystyle [a,b]\) or \(\displaystyle \int_{a}^{b}\)

Also

Trapezoidal Rule

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + 2i\Delta x) - 2\Delta x )]\).

\(\displaystyle \Delta x = \dfrac{b - a}{2n}\)

Interval: \(\displaystyle [a,b]\) or \(\displaystyle \int_{a}^{b}\)

Left Triangle Approximation

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + i\Delta x) - \Delta x )]\).

\(\displaystyle \Delta x = \dfrac{b - a}{n}\)

Interval: \(\displaystyle [a,b]\) or \(\displaystyle \int_{a}^{b}\)

Right Triangle Approximation

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + i\Delta x]\).

\(\displaystyle \Delta x = \dfrac{b - a}{n}\)

Interval: \(\displaystyle [a,b]\) or \(\displaystyle \int_{a}^{b}\)

Midpoint Triangle Approximation

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + i\Delta x) - \dfrac{1}{2}\Delta x)]\). :D Correction

\(\displaystyle \Delta x = \dfrac{b - a}{n}\)

Interval: \(\displaystyle [a,b]\) or \(\displaystyle \int_{a}^{b}\)
 
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Interesting Question

Capture.JPG
Let P(x) =x
It is clear that the area (integral from -1 to 2) is 3/2.

Question:
Can you find a different quadratic through (0,0) and (2,2)
Capture.JPG
That also has the (integral from -1 to 2) of 3/2?
 
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Correction again :eek:

Midpoint Triangle Approximation:

Area = \(\displaystyle \sum_{i = 1}^{n} \Delta x[f((a + i\Delta x) - \dfrac{1}{2}\Delta x)]\).

Always need to double check this stuff. Still pretty sure the other formulas are right.
 
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