Determine the area of the shaded region in the following figur...

Question

Determine the area of the shaded region in the following figures.

Graph showing a shaded region between two curves, with axes labeled x and y, indicating area calculation.

Accepted Answer

Hello. In this video, we are going to be finding the area enclosed by the shaded region in the given figure. Now, if we are taking a look at this given region, notice that we have the shaded region that is defined by X is equal to Y2d and the line Y is equal to 6 minus X. Now, in this case here, if we try to integrate with respect to X, we are going to get an undefined value since the rectangle of integration is going to be connected to the same curve. So for this problem, it'll be better to integrate with respect to Y, since we have a defined rightmost curve and a defined leftmost curve. So, the area function is going to be in terms of Y. And that is going to be expressed as a definite integral from A to B, of the rightmost curve, minus the leftmost curve, D Y. So, the first thing we need to do is we need to define our bounds of integration. Now, the bounds of integration are going to be in terms of why. So, we already have one function that is X in terms of Y, but we need to rewrite our second function as X in terms of Y. By solving for X, we get X is equal to 6 minus Y. Then, if we were to set both of these functions equal to each other, we will have 6 minus Y is equal to Y2. By setting this equation equal to 0, we get Y2 plus Y minus 6 is equal to 0. This is going to factor to give us Y + 3 multiplied by Y minus 2, and this is going to give us two values of Y. Y is equal to -3, and Y is equal to 2. So what this means is that the points of intersection between the two curves are going to have a value of Y, which is -3, and a value of Y which is 2. So the bounds of integration are going to be from -3 to 2. Now what we need to do is we need to identify our rightmost curve. Now, the rightmost curve in this case is going to be the red function. That function in terms of Y is going to be 6 minus Y, and that means the leftmost function is going to be X is equal to Y2d. So, if we were to plug in these values into the integral, we will have the integral from -3 to 2 of 6 minus Y minus Y2DY. Now if we were to solve this integral, we will end up with the area that is trapped between the two curves. By taking the antiderivative of each term of the integral, we will have 6 Y minus 12 Y squared, minus 1/3 Y cubed, and this is going to be evaluated from -3 to 2. By plugging in the values of the bounds, we get 6 multiplied by 2, minus 1/2 multiplied by 2 squared, minus 1/3 multiplied by 2 cubed. Then we will subtract the value of the lower boundary, which is going to be 6 multiplied by -3, minus 12 multiplied by -3 squad, minus 1/3 multiplied by -3 cubed. Now what we need to do is we need to simplify. By multiplying and combining like terms in the first group, the first group is going to simplify to be 22 divided by 3. And the second group is going to simplify to be 27 divided by 2. And taking the sum of both of these fractions together is going to give us 125 divided by 6. So this is going to be the solution for the area trapped between the two curves. The area is 125 divided by 6 square units. And with that being said, the overall solution to this problem is going to be B. 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.

Determine the area of the shaded region in the following figures.

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