Calculating the Area Under a Trig Graph

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Homework Help Overview

The discussion revolves around calculating the area under the graph of trigonometric and polynomial functions, specifically focusing on the function y = cos(x) from x = 0 to x = pi/2 and f(x) = 1 + x^2 from x = -1 to x = 2. Participants explore methods of approximation using rectangles and question the applicability of certain mathematical theorems.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using approximating rectangles to estimate areas under curves, with one participant attempting to generalize the method for n rectangles. Questions arise about the existence of formulas for summing trigonometric functions and the potential for changing function forms and intervals while maintaining equivalent areas.

Discussion Status

The conversation includes various attempts to understand the approximation methods and the limits involved in calculating areas. Some participants express uncertainty about the existence of specific formulas and the implications of changing function parameters. Guidance is offered regarding the importance of the fundamental theorem of calculus and the challenges of evaluating certain sums.

Contextual Notes

Participants are working within the constraints of approximating integrals without full access to the fundamental theorem of calculus. There is an ongoing exploration of whether changing the function and interval affects the area calculation.

Miike012
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Question: Find the area under the graph y = cos(x) from x = 0 to x = pi/2

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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I don't know of any nice formula for that sum. (That doesn't mean there isn't one.) Are you studying approximating sums for integrals but don't have the fundamental theorem of calculus yet?
 
I guess you can say that... The real question was ... estimate the area of cos(x) [0,pi/2] using 4 approximating rectangles and right endpoints..

That is easy so I wanted to try and solve for n rectangles..
 
Miike012 said:
I guess you can say that... The real question was ... estimate the area of cos(x) [0,pi/2] using 4 approximating rectangles and right endpoints..

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 = x2 if your book hasn't already done it for you. You will be able to calculate it.

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.
 
one more question...

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.
 
LCKurtz said:
I don't know of any nice formula for that sum. (That doesn't mean there isn't one.) Are you studying approximating sums for integrals but don't have the fundamental theorem of calculus yet?

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.
 
Miike012 said:
one more question...

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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