The total area between 3sin(5x) and the x-axis

  • #1
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Main Question or Discussion Point

If you were given f(x) = 3sin(5x), would it be possible to express the total area between f(x) and the x-axis between the origin and any given intercept? Basically, could you form a general equation for the total area for f(x) where x∈[0,a] and a is an x-intercept.
 

Answers and Replies

  • #2
cnh1995
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If you were given f(x) = 3sin(5x), would it be possible to express the total area between f(x) and the x-axis between the origin and any given intercept? Basically, could you form a general equation for the total area for f(x) where x∈[0,a] and a is an x-intercept.
Integration gives area under the curve. You'll get the area if you calculate ∫3sin(5x)dx with limits 0 to a.
 
  • #3
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If you were given f(x) = 3sin(5x), would it be possible to express the total area between f(x) and the x-axis between the origin and any given intercept? Basically, could you form a general equation for the total area for f(x) where x∈[0,a] and a is an x-intercept.
Yes, but to get the area, you need to account for parts of the curve that are below the x-axis. For example, on the interval ##[\pi/5, 2\pi/5]## the graph of this function is below the x-axis. On intervals such as this one, the integral gives a negative value. To to get the area between the curve and the x-axis, you would need an integral like this: ##\int_{\pi/5}^{2\pi/5}-3\sin(5x)dx##

Integration gives area under the curve. You'll get the area if you calculate ∫3sin(5x)dx with limits 0 to a.
This won't necessarily give the area. For example, ##\int_0^{2\pi/5}3\sin(5x)dx## evaluates to 0, but the area between the curve and the x-axis is 12/5.
 
  • #4
mathman
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You need to compute [itex]\int 3|sin(5x)|dx[/itex], which means dividing the domain of integration into pieces where x is a multiple of [itex]\frac{\pi}{5}[/itex].




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  • #5
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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?
 

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