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=√x, y=2x−15, and y=0

Accepted Answer

Hello. In this video, we are going to be determining the area of the region bounded by the three giving curves. Now, in order to start this problem, the first thing we're going to want to do is we're going to want to draw each of the given curves. Starting with the first curve, Y is equal to the square root of X. This curve is the standard square root function. So this is going to be the shape of Y is equal to the square root of X. Next, we will graph Y is equal to 3X minus 10. This is the equation of a line starting at -10 on the Y axis and having a slope of 3. So this is going to be the shape of Y is equal to 3X minus 10. Finally, for Y is equal to 0, Y is equal to 0 is a horizontal line that overlaps the X-axis. So, the region that is bounded by the three curves is going to be this shaded region here. Now, in order to find the area of this region, what we're going to do is we're going to come up with an integral in terms of why. So, that integral is going to be defined as the following. The area function will be in terms of Y, and this is going to equal to an integral from A to B, of our rightmost curve, minus our leftmost curve, D Y. Now, the first thing we want to do is we want to identify our bounds of integration. Now, in order to identify our bounds of integration, the first thing we need to do is we need to rewrite both of our curves in terms and X in terms of Y. So, starting with Y is equal to 3X minus 10, we are going to solve for X. We will add 10 to both sides of the equation, giving us 3 X is equal to Y + 10, and dividing by 3 will give us X's equal to 1/3 Y plus 10 divided by 3. Now, this curve here is going to be the curve of the line, that is X in terms of Y. Furthermore, this is going to be the rightmost curve for the integration. Now we will go to the 2nd curve, which is Y is equal to the square root of X. If we would solve for X in this case, all we need to do is square both sides, and we get X's equal to Y2d. This is the 2nd curve in terms of Y, and this is going to be our leftmost curve for the bounds of integration. Now, what we need to do is we need to set both of these curves equal to each other in order to solve for the bounds of integration. So that is going to be 1/3 win. Plus 10 divided by 3, and this is going to equal to Y2d. We will multiply this entire equation by 3, giving us Y + 10 is equal to 3 Y2. By setting this equation equal to 0, we get 3 Y2 minus y. -10 is equal to 0. Now what we're going to do is we're going to solve this equation by using the quadratic equation. By using the quadratic equation, we get Y is equal to 1 plus or minus the square root of 1 + 120 divided by 6. Now, this is going to simplify to give us Y is equal to 1 plus or minus 11 divided by 6, and we will end up with two values of y in this case. We get Y is equal to 1 + 11 divided by 6, which is going to simplify to 2. And we get Y is equal to 1 minus 11 divided by 6, which is going to be -10 divided by 6. Now, if we take a look at the given curve, the smallest boundary of Y that we have available for the region is located at the origin. So Y is equal to 0 at this point. That means the solution, -10 divided by 6 is a redundant solution to the bounds of integration. And that means the second point of intersection, which is located here, has a Y value of 2. So our bounds of integration are going to be from 0 to 2. Now that we have our bounds of integration, and we've identified our right and left curve, we are going to plug this into the integral in order to find the area. That is going to give us the area in terms of Y, which is equal to the integral from 0 to 2 of the rightmost curve, which is 1/3 Y plus 10/3 minus the rightmost curve, which is Y 2 D Y. All we need to do in this case is solve for the integral in order to find the area that is trapped between the three curves. Now, if we were to take the antiderivative of each of these terms, we will end up with 1/6 Y2 plus 10/3 Y minus 1/3 Y cubed, and this is going to be evaluated from 0 to 2. By plugging in the value of the bounds, we'll end up with 1/6 multiplied by 22, plus 10/3 multiplied by 2 minus 1/3 multiplied by 2 cubed. Then we are going to plug in the value of 0 into the integran or into the antiderivative, but if we plug in 0 into each term of Y, we will just end up with the overall value of 0. All we need to do now is simplify. The first term is going to simplify to be 4 divided by 6. The second term will be 20 divided by 3. And the final term will be 8 divided by 3, and combining these three fractions together is going to give us a final sum of 14 divided by 3, which is going to be the area between the three curves. 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.

Find the area of the region described in the following exercises.The region bounded by y=√x, y=2x−15, and y=0

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

The region bounded by y=√x, y=2x−15, and y=0