Find the area of the region described in the following exercis...

Question

Find the area of the region described in the following exercises.

The region bounded by y=2−|x|and y=x^2

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the area of the region bounded by Y is equal to 3 minus the absolute value of X and Y is equal to 2 X squad. So to better visualize this problem, we're going to draw a quick sketch. Of our two functions. So, we'll begin by drawing our X and Y axis. And we'll first draw the function Y is equal to 3 minus the absolute value of X. And so, this is going to be a V shape with an apex. At X equal to 0, and that apex occurs at 0.3. Next, we need to draw Y equal to 2 X2, which is going to be a problem that has An apex at the origin, and is concave up. And so, the region that we're looking for the area of, is going to be the region that is enclosed by these two functions. So, the first step is to determine where these two functions intersect. And so, we'll have to look at two different cases. We'll look at the case when X is greater than 0. And when X is greater than 0, Y equal to 3 minus the absolute value of X just becomes Y equal to 3 minus X. Since X is positive, we can drop the absolute value sign. And so to find the intersection between these two functions, we just need to set them equal to each other. So we'll have 3 minus X is equal to 2 X squared, and now we just need to solve for X. So, to do this, we're going to add X to both sides of the equation and subtract 3 from both sides of the equation. Doing so, we'll get 0 is equal to 2 X squad. Plus x minus 3, and we can actually factor this quadratic equation. So we'll have 0 is equal to the quantity of 2 X plus 3 in quantity, multiplied by the quantity of X minus 1 in quantity. So to solve for X will set each factor equal to 0. And so we'll have 2 X plus 3 is equal to 0, and solving for X, we'll get that X is equal to minus 3 divided by 2. And for the other factor, we'll have X minus 1 is equal to 0, and solving for X, we'll have that X is equal to 1. Now, we're looking at the case where X is greater than 0, so we will actually disregard the negative solution. So the solution that we're looking for is X equal to 1. We'll also look at the case when X is less than 0. So, in this case, the absolute value of X is equal to minus X, so that means our first equation, the Y is equal to 3 minus the absolute value of X, becomes 3 plus X. And this is going to have to be equal to 2 X2d to find the intersection between the two graphs. So again, we need to solve for X, so we'll subtract X from both sides of the equation and subtract 3 from both sides of the equation. In doing so, we'll get 0 as equal to 2 X squad minus X minus 3. And again, we can factor this quadratic equation. So we'll have 0 is equal to the quantity of 2 X minus 3 in quantity, multiplied by the quantity of X plus 1 in quantity. And so now we just need to set each factor equal to 0 and solve for x. We'll have 2 X minus 3 is equal to 0, and solving for X, we'll get that X is equal to 3 divided by 2. And for the other factor we'll have X plus 1 is equal to 0, and solving for X, we'll get that X is equal to minus 1. And here we're looking at X less than 0, so we're going to disregard the positive solution. So the solution that we're interested in is X equal to -1. So now that we have identified the two intersection points, We will notice now that the area between these two curves is symmetric about the Y axis. So we can focus on finding the area on one side of the Y axis and then doubling that result. So, we're gonna look at the case where X is greater than or equal to 0, so we're just looking at the right side of our graph, and then doubling that result to get the total area. So, our total area, which we'll denote as A, is going to be equal to 2, multiplied by the area of the right side of the graph, and recall that that area it's going to be given by the integral from 0 to 1. Of the quantity, and here we'll have the Function that forms the upper bound, which in our case is going to be 3 minus X, and then we're gonna subtract off the equation that forms the lower bound of the region, which is going to be 2 X squad. Inquinty DX. So now we just need to perform this integration. So integrating term by term, this is going to be equal to 2, multiplied by the quantity, and when we integrate 3 with respect to X, we get 3 X. Then we'll have minus, when we integrate X. With respect to X, we get X squad, divided by 2. They'll have minus. 2, multiplied by, and here we're integrating X2d with respect to X, which will give us X cubed divided by 3. In quantity, and this is evaluated from 0 to 1. So substituting in our limits of integration, we see that this is going to be equal to 2, multiplied by the quantity of 3, multiplied by 1, minus. 1 squared, divided by 2. Whiteness 2, multiplied by 1 cubed, divided by 3. Then minus the quantity, and here we'll substitute in our lower limit, which is going to be 3, multiplied by 0. Minus 0 squared, divided by 2. Minus 2 multiplied by 0 cubed, divided by 3 in quantity. And simplified in this expression by distributing R2 into this expression. This is going to be equal to. 6 minus 1. -4 divided by 3. Minus 0 + 0. Plus 0. And simplifying this expression, we see that our final answer is going to be equal to 11 divided by 3. So that is the answer that we're looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Find the area of the region described in the following exercises.

The region bounded by y=2−|x|and y=x^2

Key Concepts

Area Between Curves

Finding Intersection Points

Integration

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