Area Find (i) the net area and (ii) the area ...

Question

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question.

The region bounded by y = 6 cos 𝓍 and the 𝓍-axis between 𝓍 = ―π/2 and 𝓍 = π

Accepted Answer

Hello. In this video, we are asked to find the net area and the total area of the region bounded by Y is equal to 5 cosine of X, the X-axis, and we want this to be between X's equal to 0 and X is equal to 2 pi. Let's go ahead and start this problem by graphing the given information. Now, we want our boundaries to be between X's equal to 0 and X's equal to 2 pi. And one of the boundaries is going to be the X axis. Now, the function Y is equal to 5 cosine of X is the standard cosine function, but with an amplitude of 5. So, we have endpoints at 0.5 and 25. We have a minimum value located at -5, and we have X intercepts located at pi divided by 2. 0. And 3 pi divided by 2.0. And if we were to connect these dots, we will have a cowbell shape. For the cosine wave. Now, since we only want the areas that are bounded between X's equal to 0 and 2 pi and the X-axis, those bounded areas are going to be the shaded regions here. And so let's go ahead and start this problem by solving for the net area. In order to find the net area, we just want the total amount of area that is from 0 to 2 pi underneath the curve. And so in order to solve that, we can set up a definite integral. This is going to be the total area from 0 to 2 pi of the given function. And that function is going to be 5, cosine of XDX. So, let's go ahead and solve this integral in order to find the net area. We will move 5 to the front of the integral, giving us 5, multiplied by the integral from 0 to 2 of cosine of X, and the integral or the antiderivative of cosine of X is going to be sin of X. So we have 5 multiplied by sin of X, evaluated from 0 to 2 pi. Now, if we plug in the bounds, we have 5 multiplied by sin of 2 pi minus 5 multiplied by sign of 0. But the problem about this is that both of the values evaluate and simplify to be zero, and what that means is that the net area is going to be 0. So that's going to be one of the solutions for this problem. But now what we need to do is we need to find the total area. Now, in order to find the total area, we need to find the amount of area that is between or bounded in each section, and it doesn't matter whether that area is positive or negative, we only care about the total amount of area that is bounded. And so what that means is that we are going to have to find the area within these three regions and take the absolute value of those areas and sum them together. So, let's go ahead and set up an integral for the first area. Now this starts at 0 and ends at the X intercept located at pi divided by 2. So, we can set up that integral to be the integral from 0 to pi divided by 2 of our function 5 sine theta. Or sign DX. Apologies, that's going to be 5 cosine of X, because Y is equal to 5 cosine of X. But we know that the antiderivative of this already is going to be 5 sin of x evaluated from 0 to pi divided by 2. And so if we were to evaluate this and plug in the bounds, we will be left with the absolute value of 5 multiplied by 1, minus 5 multiplied by 0, which is going to give us the absolute value of 5, which simplifies to 5. So at least we know. That this first shaded region has a total area of 5. Now what we need to do is we need to repeat this process, but for the second shaded region. Now, this shaded region is from pi divided by 2 to 3 pi divided by 2. So we need to find total area. The absolute value of 5 sin of x. Evaluated from pi divided by 2 to 3 pi divided by 2. And so that is going to give us the absolute value of 5 multiplied by -1. -5 multiplied by 1. And that is going to simplify to be -5. -5, which gives us the absolute value of -10, but that is just going to simplify to be 10. So this tells us that the total area of the second shaded region is going to be 10. And finally, we just need to find the total area of the last region, that is going to be from 3 pi divided by 2 to 2 pi. So we need to find. The absolute value Of 5 sign of x. Evaluated from 3 pi divided by 2 to 2 pi. That is going to leave us with the absolute value of 5 multiplied. 5 multiplied by 0. 5 multiplied by -1. That is going to give us the absolute value of 5, which simplifies to 5, and we know that the total area underneath the third shaded region is going to be 5. Now, if we were to take the sum of the absolute area under all three of the regions, that is going to leave us with 5 + 10 + 5, which gives us a total area of 20. So the total area underneath the curve is going to be 20, but the net area is going to be zero. And with that being said, the overall solution to this problem is going to be A. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question.

The region bounded by y = 6 cos 𝓍 and the 𝓍-axis between 𝓍 = ―π/2 and 𝓍 = π

Key Concepts

Definite Integral

Net Area vs. Total Area

Graphing Trigonometric Functions

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