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 two curves with a shaded area between them, illustrating a mathematical problem involving area determination.

Accepted Answer

Hello. In this video, we are going to be finding the area enclosed by the shaded region in the given figure. Now, the figure given to us shows a parabola defined by this function here, and a line defined by 5 minus X. Now, if we are to decide how to find the area, we are going to have to find a definite integral that will calculate the area for us. Now, if we try to integrate with respect to X, we're going to end up getting an undefined value since the rectangle of integration is going to be defined by the same curve in one of these spots. So instead, it would be better to integrate with respect to Y, since we're integrating with respect to Y will always have a different curve defined at each end of the rectangle of integration. So, that means the area function is going to be in terms of Y, and that is going to be an integral from A to B, of a rightmost curve, minus the leftmost curve, D Y. So, the first thing we want to do is we want to go ahead and define the bounds of integration. Now, if we take a look at the graph here, notice that we have two intersecting points of the graph. The smallest intersecting point. is defined with the value of -3 based off the graph, and the largest intersection point has a Y value of 3. So that means that in terms of Y, we are going to be integrating from -3 to 3. Now we need to define our rightmost curve. Now, the rightmost curve in this case is going to be this red line, but we need our red line to be in terms of Y. So, what we're going to have to do is we're going to have to solve for X in the second function. So, the line is defined as Y is equal to 5 minus X, and solving for X in this case is going to give us X is equal to 5 minus Y. So this is going to be the function of the rightmost curve in terms of Y. Now, the second most curve is already in terms of Y, but let's go ahead and simplify this so that we can put it into the integral a little easier. If we were to expand the numerator, we will end up with y squared. Minus 2 Plus 1 divided by 2 and breaking this apart is going to give us y2 divided by 2 minus Y plus 1/2. This is going to be the definition of the leftmost curve for our integral. So now that we have all the pieces that we need, let's go ahead and input all of this into the integral. So we're going to have the definite integral from -3 to 3 of the rightmost curve, which is 5 minus Y minus the leftmost curve, which is Y2 divided by 2, minus Y plus 12 D Y. Now, all we need to do is we need to solve this integral in order to find the area that is trapped between the two curves. The first thing we're going to do is we're going to distribute that negative sign in the integrand. That will leave us with the integral from -3 to 3 of 5 minus Y minus 12 Y2 + Y minus 12 D Y. Then if we were to combine like terms, we will have the integral from -3 to 3 of 9 halves minus 1/2 Y 2 D Y. Now, if we were to take the antiderivative of each term within the integrand, we will have 9 halves Y minus 1/6 Y cubed, and this will be evaluated from -3 to 3. All we need to do now is plug in the values of the boundaries. That will leave us with 9 halves multiplied by 3 minus 16 multiplied by 3 cubed. Then we will subtract the value of the lower boundary, which is 9 halves multiplied by -3 minus 1/6 multiplied by -3 quantity cubed. All we need to do now is simplify. By multiplying and combining like terms together in both groups, we will have 27 minus 9, and this is going to simplify to give us 18. So, what this means is that the total area that is trapped between two curves is going to be 18 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.

Key Concepts

Area Between Curves

Integration

Finding Points of Intersection

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