Finding the Area Beneath Trigonometric Curves (1 Viewer)

[FDownes]

I know the solution to this particular question, but I'm not too sure of the reasoning behind it. The question asks;

Two curves, y = (sqrt3)cosx and y = sinx, are drawn on a graph. The first two intersections to the right of the y-axis are labelled A and B.

a) Solve the equation (sqrt3)cosx = sin x to find the x-coordinates of A and B.

b) Find the area contained between the graphs of y = (sqrt3)cosx and y = sinx between the points A and B.

My problem is not in finding the solution, but in how the solution is obtained. For some reason, to find the area, you simply find the integral of sinx - (sqrt3)cosx between the points B (x = 4pi/3) and A (x = pi/3). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and pi/2 (i.e. the integral of sinx - (sqrt3)cosx between x = pi/2 and x = pi/3) and another to find the area betwee B and pi/2 (i.e. the integral of (sqrt3)cosx - sinx between x = 4pi/3 and x = pi/2). I would have thought that since part of the graph is below the x-axis then the area would be partly negative and therefore would counterract some of the area enclosed above the x-axis.

Why is it that you can find the area, which is 4 units², using only a single integral, instead of two?

 

[shaon0]

I know the solution to this particular question, but I'm not too sure of the reasoning behind it. The question asks;

Two curves, y = (sqrt3)cosx and y = sinx, are drawn on a graph. The first two intersections to the right of the y-axis are labelled A and B.

a) Solve the equation (sqrt3)cosx = sin x to find the x-coordinates of A and B.

b) Find the area contained between the graphs of y = (sqrt3)cosx and y = sinx between the points A and B.

My problem is not in finding the solution, but in how the solution is obtained. For some reason, to find the area, you simply find the integral of sinx - (sqrt3)cosx between the points B (x = 4pi/3) and A (x = pi/3). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and pi/2 (i.e. the integral of sinx - (sqrt3)cosx between x = pi/2 and x = pi/3) and another to find the area betwee B and pi/2 (i.e. the integral of (sqrt3)cosx - sinx between x = 4pi/3 and x = pi/2). I would have thought that since part of the graph is below the x-axis then the area would be partly negative and therefore would counterract some of the area enclosed above the x-axis.

Why is it that you can find the area, which is 4 units², using only a single integral, instead of two?

Just solve:
a) sqrt(3)cosx=sinx
1/sqrt(3)=tanx (Or you could use auxillary methods)
Solve from here.
b)then integrate with limits

 

[lolokay]

sin[x] - rt3cos[x] = 2sin[x - pi/3]
= 0 when x = pi/3, 4pi/3 etc.
so sin[x] - rt3cos[x] is always positive between those limits, which is why the single integral gives you the area

I'm pretty sure that as long as the area between the 2 curves is in the one section (ie they don't intersect between the limits) then just integrating the difference of the 2 as a single integral will always be fine - since the difference between the 2 can either only be positive, or be only negative (in which case you take the absolute value anyway)

 

[shaon0]

sin[x] - rt3cos[x] = 2sin[x - pi/3]
= 0 when x = pi/3, 4pi/3 etc.
so sin[x] - rt3cos[x] is always positive between those limits, which is why the single integral gives you the area

I'm pretty sure that as long as the area between the 2 curves is in the one section (ie they don't intersect between the limits) then just integrating the difference of the 2 as a single integral will always be fine - since the difference between the 2 can either only be positive, or be only negative (in which case you take the absolute value anyway)

up this late lolokay? its 1:23am

 

[lolokay]

I've got no more school

 

[shaon0]

I've got no more school

same, just finished my physics exam