Question Video: Finding the Area of the Region Bounded by Two Trigonometric Functions Mathematics • Higher Education

Find the area of the region enclosed by the curves 𝑦 = 16 cos π‘₯ and 𝑦 = 2 secΒ² π‘₯ for π‘₯ between βˆ’πœ‹/3 and πœ‹/3.

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Video Transcript

Find the area of the region enclosed by the curves 𝑦 is equal to 16 cos of π‘₯ and 𝑦 is equal to two sec squared of π‘₯ for π‘₯ between negative πœ‹ by three and πœ‹ by three.

In this question, we’re asked to find the area of a region enclosed by two curves, 𝑦 is equal to 16 cos of π‘₯ and 𝑦 is equal to two sec squared π‘₯, and between two vertical lines, π‘₯ is equal to negative πœ‹ by three and π‘₯ is equal to πœ‹ by three. And whenever we’re asked to find an area of a region enclosed by curves, we should always sketch a diagram, because it can help us determine the easiest way to find this area.

Before we start sketching our curves, we should note that our values of π‘₯ will range between negative πœ‹ by three and πœ‹ by three. So we only need to sketch these curves for these values of π‘₯. So let’s start by sketching 𝑦 is equal to 16 cos of π‘₯ between π‘₯ is negative πœ‹ by three and πœ‹ by three. We can do this by noticing this is just the graph 𝑦 is equal to cos of π‘₯ stretched by a factor of 16 in the vertical direction. So this is just a vertical stretch of the cosine function. Its graph looks something like the following.

On the same axis, we need to sketch the curve 𝑦 is equal to two sec squared of π‘₯. And we can see that this will be a vertical stretch of factor two of the secant function. But remember, we’re sketching this on the same axis we’re sketching 𝑦 is equal to 16 cos of π‘₯. So we need to make sure we get our scales correct. One way of doing this is to consider a few points on our graph. First, we can find the 𝑦-intercept of 𝑦 is equal to 16 cos of π‘₯ by substituting π‘₯ is equal to zero into the function. cos of zero is one, so the 𝑦-intercept is 16. We can then find the 𝑦-intercept of our other curve. We substitute π‘₯ is equal to zero to get two sec squared of zero. The sec of zero is one, so the 𝑦-intercept is at two.

So we can add this point onto our diagram. And now we’re almost ready to sketch our curve. We know the shape of this function. However, it will be useful to find the endpoints of this curve. We’ll do this by substituting π‘₯ is negative πœ‹ by three and π‘₯ is πœ‹ by three into the function. We can do both of these at the same time by noticing that the secant function is an even function. Therefore, two sec squared of negative πœ‹ by three is equal to sec squared of πœ‹ by three. We’ll rewrite this by noting the sec of πœ‹ by three is one divided by the cos of πœ‹ by three. So we get two divided by cos squared of πœ‹ by three.

Next, the cos of πœ‹ by three is one-half. So we get two divided by one-half squared, which we can evaluate is equal to eight. However, at this point, we can notice something interesting. The cosine function is also an even function. And 16 cos of positive or negative πœ‹ by three is equal to eight. So the 𝑦-coordinates of the endpoints of our curve 𝑦 is equal to 16 cos of π‘₯ is eight. So the endpoints of the two curves meet up, which give us a sketch which looks like the following. We can now shade in the region we need to determine the area of. We can now see that this region is bounded above by the curve 𝑦 is equal to 16 cos of π‘₯ and below by 𝑦 is equal to two sec squared of π‘₯. And it’s bounded between two vertical lines: π‘₯ is negative πœ‹ by three and π‘₯ is equal to πœ‹ by three.

We can then recall if we have two integrable functions 𝑓 and 𝑔 and 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ on a closed interval from π‘Ž to 𝑏, then the area between the curves 𝑦 is equal to 𝑓 of π‘₯ and 𝑦 is equal to 𝑔 of π‘₯ and the vertical lines π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏 is given by the definite integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯. And we can see this is true of the region we need to find the area of. 𝑓 of π‘₯ is the upper function, 16 cos of π‘₯. 𝑔 of π‘₯ is the lower function, two sec squared of π‘₯. π‘Ž is the lower bound of π‘₯, negative πœ‹ by three. And 𝑏 is the upper bound of π‘₯, πœ‹ by three.

Finally, it’s also worth pointing out we know that our two functions are integrable because they’re continuous on this interval. Therefore, substituting this information into the integral, we get the area of the shaded region is the integral from negative πœ‹ by three to πœ‹ by three of 16 cos of π‘₯ minus two sec squared of π‘₯ with respect to π‘₯. So we now need to evaluate this integral. We can do this term by term by recalling two integral results.

First, we recall the integral of the cos of π‘₯ with respect to π‘₯ is the sin of π‘₯ plus the constant of integration 𝐢. Second, we recall the integral of the sec squared of π‘₯ with respect to π‘₯ is the tan of π‘₯ plus the constant of integration 𝐢. Of course, in this case, we’re working with definite integrals, so we don’t need to include the constants of integration. Therefore, 16 sin of π‘₯ is an antiderivative of 16 cos of π‘₯ and negative two tan of π‘₯ is an antiderivative of negative two sec squared of π‘₯. This gives us 16 sin of π‘₯ minus two tan of π‘₯ evaluated at the limits of integration π‘₯ is negative πœ‹ by three and π‘₯ is πœ‹ by three.

We now need to find the difference of the antiderivative evaluated at the two limits of integration. And this is given by 16 sin of πœ‹ by three minus two tan of πœ‹ by three minus 16 sin of negative πœ‹ by three minus two tan of negative πœ‹ by three. And we could just evaluate this expression here. However, we can simplify this by noticing both sine and tangent are odd functions. Therefore, sin of negative πœ‹ by three is negative one times the sin of πœ‹ by three. Similarly, the tan of negative πœ‹ by three is negative one multiplied by the tan of πœ‹ by three.

If we then distribute the negative over our parentheses, we can see that we get another term of positive 16πœ‹ by three and another term of negative two tan of πœ‹ by three. Therefore, this expression is equal to 32 sin of πœ‹ by three minus four tan of πœ‹ by three. And now we can evaluate this expression by first recalling the sin of πœ‹ by three is root three over two. So 32 sin πœ‹ by three is root three over two multiplied by 32. We can cancel the shared factor of two in the numerator and denominator to get 16 root three. Similarly, we can recall the tan of πœ‹ by three is equal to root three. If we multiply both sides of this equation through by negative four, we get negative four tan of πœ‹ by three is negative four root three. We can then substitute both of these values into our expression.

We’ll first clear some space. Then, we substitute these expressions. We get 16 root three minus four root three, which we can calculate is 12 root three, which is our final answer. And it’s worth noting this represents an area, so we could say that this is 12 root three square units. Therefore, we were able to show the area of the region enclosed by the curves 𝑦 is equal to 16 cos of π‘₯ and 𝑦 is equal to two sec squared of π‘₯ for values of π‘₯ between negative πœ‹ by three and πœ‹ by three is 12 root three.

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