Sketch the region in the first quadrant enclosed by y=6/x, y=2x, and y=.5x. Decide whether to integrate with respect to x or y. Then find the area of the region.
need final area as answer please, and step would be helpful also if you can post.
sketch it of course
where does y = 2x hit y=6/x?
2 x = 6/x
x^2 = 3
x = sqrt 3
then y = 2 sqrt 3
so
from x = 0 to x = sqrt 3 integrate y = (2 x-.5 x) = 1.5 x
1.5 x^2/2
.75 ( 3-0) = 2.25
then where does y = .5 x hit y = 6/x?
.5 = 6/x^2
x^2 = 12
x = 2 sqrt 3
so integrate y = (6/x - .5 x) from x = sqt 3 to x = 2 sqrt 3
(6 ln x - x^2/4)
6 ln (2 sqrt 3 - sqrt 3) - .25(12-3)
2.62
so 2.25 + 2.62
Nick, with a little effort on your part to complete the solution I gave you and some small adjustment for this question, you could have saved
the duplication of this solution by Damon.
Oh my :(
To sketch the region in the first quadrant enclosed by the curves y = 6/x, y = 2x, and y = 0.5x, we can start by graphing each of these equations separately.
First, let's look at the equation y = 6/x. To plot this curve, we can choose some x-values and calculate the corresponding y-values using the equation. For example, if we select x = 1, then y = 6/1 = 6. We can do this for a few more x-values, such as x = 2, 3, and so on. Here's a table showing some values:
x | y = 6/x
------------
1 | 6
2 | 3
3 | 2
4 | 1.5
5 | 1.2
Plotting these points and connecting them, we get a curve that decreases as x increases. This curve will pass through the negative y-axis and approach the positive x-axis.
Next, let's look at y = 2x. This is a linear equation with a slope of 2, meaning for every unit increase in x, y increases by 2 units. We can plot a few points to get an idea of this line. For example, if x = 1, then y = 2(1) = 2. We can do this for a few more x-values:
x | y = 2x
--------------
1 | 2
2 | 4
3 | 6
4 | 8
5 | 10
Plotting these points and connecting them, we get a line that starts at the origin and continues upwards with a slope of 2.
Finally, let's look at y = 0.5x. This is another linear equation, but with a slope of 0.5. Similarly, we can plot a few points:
x | y = 0.5x
--------------
1 | 0.5
2 | 1
3 | 1.5
4 | 2
5 | 2.5
Plotting these points and connecting them, we get a line that starts at the origin and continues upwards with a smaller slope of 0.5.
Now, to find the enclosed region, we can shade the area between these three curves in the first quadrant.
| /
| /
|-------/-----
| /
| / (enclosed region)
To determine whether to integrate with respect to x or y, we need to consider the shape of the region. Looking at the sketch, we can see that the region is bounded by curves in terms of y: y = 2x (the upper curve) and y = 0.5x (the lower curve).
Therefore, we will integrate with respect to y.
To find the area of the region, we need to evaluate the integral from the lower curve to the upper curve with respect to y.
Using the curves y = 0.5x and y = 2x, we can find the x-values at which they intersect. Setting these two equations equal to each other: 0.5x = 2x, we solve for x:
0.5x = 2x
0.5x - 2x = 0
-1.5x = 0
x = 0
So, the curves intersect at x = 0. Since we are looking for the region in the first quadrant, we can disregard this intersection.
Now, the area can be found by evaluating the integral of the upper curve (y = 2x) minus the integral of the lower curve (y = 0.5x) over the appropriate y-range.
∫[lower bound, upper bound] (2x - 0.5x) dy
We need to find the limits of integration, which are the y-values where the curves intersect. Since the curves are linear, they intersect at y = 0. To find the upper bound, we set y = 2x and solve for x:
y = 2x
0 = 2x - y
2x = y
x = y/2
So, the upper bound for integration is x = y/2.
Now, we can evaluate the integral:
∫[0, y_upper_bound] (2x - 0.5x) dy
∫[0, y_upper_bound] (1.5x) dy
Using the upper bound x = y/2, we can substitute this into the integral:
∫[0, y_upper_bound] (1.5 * y/2) dy
∫[0, y_upper_bound] (0.75y) dy
Integrating with respect to y:
[0.75 * (1/2) * y^2] evaluated from 0 to y_upper_bound
0.375 * y^2 evaluated from 0 to y_upper_bound
Now, we substitute the upper bound y_upper_bound into the expression and subtract the result of substituting 0:
0.375 * y_upper_bound^2 - 0.375 * 0^2
0.375 * y_upper_bound^2 - 0
Since we're looking for the final area, we can just square the upper bound
0.375 * (y_upper_bound)^2
Now, we need to find the upper bound by plugging in y/2:
0.375 * (y/2)^2
0.375 * y^2 / 4
0.09375 * y^2
So, the area of the region enclosed by the curves y = 6/x, y = 2x, and y = 0.5x in the first quadrant is 0.09375 * y^2.
To calculate the numeric value of the area, we would need to know the upper bound value of y.