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?
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?
Last edited: