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Forums
Mathematics
Calculus
The total area between 3sin(5x) and the x-axis
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[QUOTE="Saracen Rue, post: 5495635, member: 521193"] I have done some thinking about this an I think I've figured out a way to do it. The graph intersects the x-axis every n*pi/5 interval. (i.e. 0*pi/5 = 0, 1*pi/5 = pi/5, 2*pi/5, 3*pi/5, etc) Due to symmetry in the graph, we also know the area between the positive and negative sections of the graph will also be the same (i.e. the area between (0,0) to (pi/5, 0) and (pi/5, 0) to (2pi/5, 0) is the same, however integrating for the latter area results in a negative. However, this means if we just calculate the area over just (0,0) to (pi/5, 0) and multiply it by the number of areas the graph has over the given domain, we should get the total area. The domain is [0, a], where a is an x-intercept. This means a = n*pi/5. As n also equals the number of areas up to that point (i.e. at 7pi/5, there have been 7 areas above/below the x-axis) a = n*pi/5 n = 5a/pi As discussed earlier, the total area should equal the area over the domain [0, pi/5] multiplied by the number of peaks (n) over the total domain. Total area = n * integration(5pi/3, 0, 3sin(5x)) = 5a/pi * integration(5pi/3, 0, 3sin(5x)) Is this a correct answer? [/QUOTE]
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Forums
Mathematics
Calculus
The total area between 3sin(5x) and the x-axis
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