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# Sketch the region in the xy-plane defined by the inequalities $x - 2y^2 \ge 0$ , $1 - x - \mid y \mid \ge 0$ and find its area.

## $\frac{7}{12}$

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Applications of Integration

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### Video Transcript

we want to sketch the region defined by these two inequalities and then determine the area of that region. Okay, so we're gonna do a sketch in our X Y plane. As usual, we have your ex and we have our y axes. Okay, Uh, so with these inequalities, usually what we'd like to do is to sulfur. Why as a function of X. But however, because of this absolute value of why, over here, um, we're actually going to solve for X in terms of why, For both of these inequalities, this is going to give us X is greater than or equal to two. Why squared and in green, we're going to have X is less than or equal to one, minus the absolute value of why. Okay, so now what we have here are two ah, functions of why so when we put these, we're going to kind of think of the graph is being sideways. So let's put the blue one first. Um X. So let's take a look at, um, X is equal to two. I squared. So thinking of X as so we see that this is going to be a problem. Um, but It's going to be a sideways problem just by thinking of this with her head turned sideways. And then what we're interested in is ex being greater than or equal to two y squared. So that's actually going to be the inside region here. Now let's take a look at the green one. Ah, we have X less than or equal to one minus y squared. So first, let's plot. Why equals one minus absolute? Why? So we know that when why is zero We have our point here at one. And then we're subtracting an absolute value, which means that as we go away from Y equals zero, our ex values will decrease. So we are interested in X being less than or equal to one minus absolute way, which is going to give us this region over here. So in fact, we see that there is an enclosed region, or rather, a region, uh, which overlaps between these two inequalities and let's color it in red. So that's the area that we want to determine. So it's going to be more instructive to do this as an integral of why, so that we can integrate. Um, we see that this function is even or symmetric across this X axis. So if we just determined one of these areas, we can double it to determine the total area. Okay, let's put this into equations. Area is going to be equal to two times the area of for example, let's take the top region. So we know that we're going to be integrating from here. Why equals zero to here, which is why equals? We have to determine the intersection point. So let's do that. We want to find out what points are two ex functions which are functions of way. Uh, at what point do they intersect? Remember, our first function was X equals two y squared, and the second function is X equals one minus absolute way. So we want to determine when these two functions intersect. Remember, we are taking why greater than zero. So this is going to tell us that this second function is just going to be one minus y. So we wanted a cell for two y squared is equal to one minus way or bringing everything to one side to y squared plus y minus one equals zero so we can use the quadratic formula we're going to give. I'd get to values for why? Why is going to be either negative one or a positive half? But since we're interested in why being greater than or equal to zero, we're going to take y to be 1/2. So what does this tell us? This tells us if we look back to the graph, this upper value for why is actually why equals 1/2. So are integral is gonna be going from why equals zero upto y equals half and it's going to be a d. Y integral. Since we're integrating in the UAE variable No, we have to use our top function minus their bottom function. So what is our top function here? Well, thinking of this again with her head turned sideways, this green function is on top on by on top. I mean, there it takes hire ex values and the bottom function will be the blue one since it takes lower since it takes lower X values. So what we're going to do is take our tough function, which is green, and we subtract our bottom function, which is a blue. So we're going to do one minus y minus two y squared Notice here. I didn't put in the absolute value since we're only considering X being positive. Okay, so now we just carry out this integral. So is equal to two anti derivative Here is why minus half. Why squared minus to over three. Why? To the three. And we're going from 0 to 1/2. So I plug in our limits and we see what we get. So we have a two out here plugging in like was half. We get half minus half times one over four, minus 2/3 times half cubed is one over eights. And then after simplifying everything, we get our final answer off. Seven over 12. As the area of the enclosed region of the total enclosed region.

University of Toronto

#### Topics

Applications of Integration

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