- #1

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Solution:

A = Lim ( ∏/(2n) * Ʃ cos( ∏i/(2n)) = ? Start: i = 0 and End: n = n

n → ∞

Just like there is a theorem for adding consecutive numbers.... n(n + 1)/2..

Is there one for trig functions????

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- Thread starter Miike012
- Start date

- #1

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- 0

Solution:

A = Lim ( ∏/(2n) * Ʃ cos( ∏i/(2n)) = ? Start: i = 0 and End: n = n

n → ∞

Just like there is a theorem for adding consecutive numbers.... n(n + 1)/2..

Is there one for trig functions????

- #2

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- #3

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That is easy so I wanted to try and solve for n rectangles..

- #4

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That is easy so I wanted to try and solve for n rectangles..

That's good that you found it so easy. That means you get the idea and that is what counts. There is a reason calculus books typically only do the limit thing for parabolas. Try that if for y = x

Trying it for most functions will leave you with a sum that you can't evaluate in closed form, such as you have just experienced. That is why the fundamental theorem of calculus is so important.

- #5

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Estimate the area under the graph of f(x) = 1 + x^2 from x = -1 to x = 2 using 6 rectangles and right end point.

Question:

Can I change the equation from 1 + x^2 to 1 + (x - 1)^2 and change the interval to x = 0 to x = 3 ??

This seems logical because technically it would be the same area.. and it is easier for me to break up into 6 rectangles.

- #6

Dick

Science Advisor

Homework Helper

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Just for the record, there are formulas like that. See http://en.wikipedia.org/wiki/List_of_trigonometric_identities Look under "Other sums of trigonometric functions". You can derive them by summing the geometric series exp(i*a*k) and splitting into real and imaginary parts.

- #7

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Estimate the area under the graph of f(x) = 1 + x^2 from x = -1 to x = 2 using 6 rectangles and right end point.

Question:

Can I change the equation from 1 + x^2 to 1 + (x - 1)^2 and change the interval to x = 0 to x = 3 ??

This seems logical because technically it would be the same area.. and it is easier for me to break up into 6 rectangles.

Yes you could, but I don't see why it is any easier.

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