Finding Right Endpoints without Seeing a Graph

[Jason76]

Dealing with endpoints of vertical rectangles on a graph (primitive integration)

\(\displaystyle f(x) = 4 \cos(x)\)

Using formulas:

\(\displaystyle [a, b]\) - x interval

\(\displaystyle n\) - number of intervals

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

\(\displaystyle \sum (\Delta x) [f(a + (i)(\Delta x)]\) - Formula for right endpoints

Given:

Interval: \(\displaystyle [0,\dfrac{\pi}{2}]\)

\(\displaystyle n = 4\)

\(\displaystyle \Delta x = \dfrac{\dfrac{\pi}{2} - 0}{4} = \dfrac{\pi}{8}\)

Problem:

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(0 + (1)(\dfrac{\pi}{8})] + [f(0 + (2)(\dfrac{\pi}{8})] + [f(0 + (3)(\dfrac{\pi}{8})] + [f(0 + (4)(\dfrac{\pi}{8})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[f(\dfrac{\pi}{8})] + [f(\dfrac{\pi}{4})] + [f(\dfrac{3\pi}{8})] + [f(\dfrac{\pi}{2})]]\)

\(\displaystyle \sum (\dfrac{\pi}{8}) [[4 \cos(\dfrac{\pi}{8})] + [4 \cos(\dfrac{\pi}{4})] + [4 \cos(\dfrac{3\pi}{8})] + [4 \cos(\dfrac{\pi}{2})]]\) Any idea on right values?

 

[stapel]

If you're asking about evaluating the cosine terms exactly, this is something you would have learned back in trig or pre-calculus. Try looking up "double-" and "half-angle identities", and using this information to convert the pi-over-8's into pi-over-4's. :wink:

 

[Jason76]

If you're asking about evaluating the cosine terms exactly, this is something you would have learned back in trig or pre-calculus. Try looking up "double-" and "half-angle identities", and using this information to convert the pi-over-8's into pi-over-4's. :wink:

I can find exact values with a calculator also. But the values given are often strange numbers.