Finding the Area Beneath Trigonometric Curves

**Flay** · September 22nd 2008, 11:57 PM

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

Two curves, $y = \sqrt{3}cosx$ 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 $\sqrt{3}cosx$ = $sinx$ to find the x-coordinates of A and B.

b) Find the area contained between the graphs of $y = \sqrt{3}cosx$ 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 $sinx - {sqrt}3cosx$ between the points B ( $x = \frac{4\pi}{3}$ ) and A ( $x = \frac{\pi}{3}$ ). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and $\frac{\pi}{2}$ (i.e. the integral of $sinx - \sqrt{3}cosx$ between $x = \frac{\pi}{2}$ and $x = \frac{\pi}{3}$ ) and another to find the area betwee B and $\frac{\pi}{2}$ (i.e. the integral of $\sqrt{3}cosx - sinx$ between $x = \frac{4\pi}{3}$ and $x = \frac{\pi}{2}$ ). 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 $4 units^2$ , using only a single integral, instead of two?

**mr fantastic** · September 23rd 2008, 02:52 AM

Originally Posted by Flay

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

Two curves, $y = \sqrt{3}cosx$ 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 $\sqrt{3}cosx$ = $sinx$ to find the x-coordinates of A and B.

b) Find the area contained between the graphs of $y = \sqrt{3}cosx$ 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 $sinx - {sqrt}3cosx$ between the points B ( $x = \frac{4\pi}{3}$ ) and A ( $x = \frac{\pi}{3}$ ). I would have thought that you would need to find two seperate integrals, one to find the area enclosed between A and $\frac{\pi}{2}$ (i.e. the integral of $sinx - \sqrt{3}cosx$ between $x = \frac{\pi}{2}$ and $x = \frac{\pi}{3}$ ) and another to find the area betwee B and $\frac{\pi}{2}$ (i.e. the integral of $\sqrt{3}cosx - sinx$ between $x = \frac{4\pi}{3}$ and $x = \frac{\pi}{2}$ ). 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 $4 units^2$ , using only a single integral, instead of two?

If f(x) > g(x) over [a, b] then the area bounded by the curves y = f(x) and y = g(x) between x = a and x = b is equal to $\int_a^b f(x) - g(x) \, dx$ .

Since $\sin x > \sqrt{3} \cos x$ over $\left[ \frac{\pi}{3}, \, \frac{4 \pi}{3} \right]$ the bounded area is $\int_{\pi/3}^{4 \pi/3} \sin x - \sqrt{3} \cos x \, dx$ .

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