Calculating the Area Under a Trig Graph

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SUMMARY

The discussion focuses on calculating the area under the graph of the function y = cos(x) from x = 0 to x = π/2 using Riemann sums. The initial approach involves using the limit of a sum with n rectangles, but the conversation highlights the challenges of evaluating such sums without the fundamental theorem of calculus. Additionally, participants explore estimating areas under other functions, such as f(x) = 1 + x², and discuss the implications of changing the function and interval while maintaining the same area. The importance of understanding Riemann sums and the fundamental theorem of calculus is emphasized throughout the discussion.

PREREQUISITES
  • Understanding of Riemann sums and their application in calculus
  • Familiarity with the fundamental theorem of calculus
  • Basic knowledge of trigonometric functions, specifically cosine
  • Ability to manipulate algebraic expressions and functions
NEXT STEPS
  • Learn how to apply Riemann sums to various functions
  • Study the fundamental theorem of calculus in detail
  • Explore trigonometric identities and their applications in calculus
  • Practice estimating areas under curves using different methods, including rectangles and trapezoids
USEFUL FOR

Students studying calculus, educators teaching integration techniques, and anyone interested in understanding the application of Riemann sums and the fundamental theorem of calculus in estimating areas under curves.

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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