# integrating sine functions

• Oct 20th 2008, 05:13 AM
Mouseman
integrating sine functions
Posted, it here as the homework has no help was given in the other thread, and it has become a matter which quite urgently must be solved.

I am to determine the area between these two curves. Although I am familiar to the process I fail to recognize and grasp what one should do to these specific types of functions. If possible could someone assist me and post a solution along with work so I can process it.

To the question.

We are to find the area between these two curves
y = 8.5sin(0.32x-0.12)+0.56
y = 4.7sin(0.42x+0.5)-4.3

From x=-10 to x=10

• Oct 20th 2008, 06:49 AM
ticbol
A diagram would be very helpful, but I don't know how to graph in graphing calculators or computers so I cannot see the actual, precise figure. But I can "see" that the
dA = (difference in y's)*dx

As for the areas that are "negatives", if there are, I cannot see those without the figure or a sketch. Although I can do the graphs roughly by sketching them on paper here, I will not do that because the idea is to show you only the general way to do your question here.

dA is vertical. height is (y2 -y1). width is dx.

dA = [(8.5sin(0.32x -0.12) +0.56) -(4.7sin(0.42x +0.5) -4.3)]*dx
dA = [8.5sin(0.32x -0.12) -4.7sin(0.42x +0.5) +4.86)]*dx

A = INT.(-10 to 10)[8.5sin(0.32x -0.12) -4.7sin(0.42x +0.5) +4.86)]*dx

A = [(8.5/0.32){-cos(0.32x -0.12)} -(4.7/0.42){-cos(0.42x +0.5)} +(4.86)x]|(-10 to 10)

A = [-26.5625cos(0.32(10) -0.12) +11.1905cos(0.42(10) +0.5) +4.86(10)]
-[-26.5625cos(0.32(-10) -0.12) +11.1905cos(0.42(-10) +0.5) +4.86(-10)]

A = 74.9735 -(-31.9498) = 106.9233 sq.units -------answer.
• Oct 20th 2008, 01:55 PM
Mouseman
I thank you for your response, however the functions do in fact cross the x axis. Is there any specific method to find that area too? Or should one derive a function by means of stationary points etc.?
• Oct 20th 2008, 02:19 PM
ticbol
Quote:

Originally Posted by Mouseman
I thank you for your response, however the functions do in fact cross the x axis. Is there any specific method to find that area too? Or should one derive a function by means of stationary points etc.?

Deriving a single function from the two given functions will give another sine curve only. You then cannot get the area between the two given functions.

By "negative" areas, I meant the areas if the upper sine curve goes below the lower sine curve.
Are there intervals, in the given domain, where the [y2 = 8.5sin(.....] curve goes below the [y1 = 4.7sin(.....] curve? Because in those intervals, (y2 -y1) will be negative....so the dA will be "negative".

As for the two given curves going down the x-axis, the (y2 -y1) height of the dA will remain positive as long as y2 is above y1 even if both curves or one curve is below the x-axis.
• Oct 20th 2008, 02:22 PM
Mouseman
I thank you again to clear up the confusion evoked in my response(Rofl). There are no such areas, and I feel confident I could apply the method proposed in other exercises.