14–25. {Use of Tech} Areas of regions Determine the area of th...

Question

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region in the first quadrant bounded by y = x/6 and y = 1−|x/2−1|

Accepted Answer

Welcome back, everyone. In this problem, we want to calculate the area in the first quadrant bounded by Y equals 1/3 of X and Y equals 2 minus the absolute value of X. A says it's 5 halves square units, B 3 halves, C 2/3, and D 25 square units. Now if we're going to find the area bounded by both of these equations in the first quadrant, first of all, let's find the points of intersection. Now, since. We're in the first quadrant. OK. Then that tells us both the values of X and Y are going to be greater than or equal to 0, OK? So I is greater than or equal to 0 and X is greater than or equal to 0, which would mean then for our equation Y equals 2 minus the absolute value of X, OK. Since the absolute value of X is going to be positive in the first quadrant, our equation is going to become Y equals 2 minus X, OK? So that's really, really helpful to help us find the points of intersection. So At the points of intersection. OK. Then we can equate both, which then that means we can say a third of X is going to be equal to 2 minus X. Now if we go ahead and solve for X at this point, we can multiply both sides by 3 to get X equal to 6 minus 3 X. Which means then that 4 X, if we add 3 X to both sides, 4 X equals 6, so X will be equal to 3 halves, OK? In other words, or this is going to be our single point of intersection. So now if we think about what the graph is going to look like. I do a quick sketch of this graph here and let me move this over since we know we're only using the first quadrant, OK. Then for starters, for our equation Y equals 2 minus X, we know that it will intersect the Y axis at 2 and it's going to have a negative slope. So it maybe looks something like that, OK. Actually, let me highlight that in a different color. So let's put that in red. OK, Y equals 2 minus X. So we know that might look something like this. On the other hand, for our graph of Y equals 3 of X, that will have a a positive uh slope and it will pass through the origin. So that may look something like this, OK? And we know that both of them intersect at the point where X equals 3 houses. So, the area, OK, that we're looking for then must be this area in the first quadrant, this area shaded in black. Now looking at our curve, we can, looking at our graph rather, the upper curve is going to be Y equals 2 minus X and the lower curve is Y equals 3 of X on the interval from 0 to 3 halves. So in that case. That means the area between the curves, OK. Can be found By finding the definite integral between the limits of 0 and 3 halves of the upper curve 2 minus X minus the lower curve a third of X all with respect to X. No we can solve. First, let's simplify the integran. Now when we simplify that here, OK, we're gonna have 2. Minus X minus 1/3 of X, which is gonna give us 2 minus for 1/3 of X, all with respect to X. And we can integrate each of these separately. Now the integral of 2 is 2 X, OK? And the integral of 4/3 of X will be 4 X squared divided by 6. OK. So that is gonna be between the bones of 0 and 3 halves, and we could simplify our second term further to make it be 2/3 of X squared. Now we can substitute our upper and lower limits. For our upper limit. OK. This is going to be 2 multiplied by 3 halves, OK, minus 2/3 of 3 halves squared. And for our lower limit, if we substitute 0 into both terms, it will simply leave us with 0. So we're focused on solving here. Now 2 multiplied by 3 halves equals 3. And uh 2/3 multiplied by 3 halves, that's 2/3 multiplied by 9/4 which is equal to 3 halves, OK? or it is equal to 3 halves here when we when we multiply them and 3 minus 3 halves equals 3 halves. Therefore, the area bounded by our curves will be equal to 3 halves square units. When we look back on our answer choices, that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

14–25. {Use of Tech} Areas of regions Determine the area of the given region.

The region in the first quadrant bounded by y = x/6 and y = 1−|x/2−1|

Key Concepts

Understanding Piecewise and Absolute Value Functions

Finding Points of Intersection

Definite Integration to Find Area Between Curves

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