Sketch the region in the first quadrant enclosed by y=7/x, y=2x, and y=x/2. Decide whether to integrate with respect to x or y. Then find the area of the region.
It doesn't really matter which variable is used, since the boundary changes in either direction.
Cleverly enough, the hyperbola intersects the two lines at (√(7/2),√14) and (√14,√(7/2).
So,
a = ∫[0,√(7/2)] (2x)-(x/2) dx
+ ∫[√(7/2),√14] (7/x)-(x/2) dx
or, along y,
a = ∫[0,√(7/2)] (2y)-(y/2) dy
+ ∫[0,√(7/2)] (7/y)-(y/2) dy
To sketch the region in the first quadrant enclosed by the three curves: y = 7/x, y = 2x, and y = x/2, we first need to visualize each curve.
1. y = 7/x:
- This curve represents a hyperbola that passes through the points (1,7), (2,3.5), (3,2.33), and so on.
- As x increases, y decreases, and vice versa.
2. y = 2x:
- This curve represents a straight line passing through the origin (0,0).
- As x increases, y increases in a linear fashion.
3. y = x/2:
- This curve represents a line with a y-intercept at (0,0) and a slope of 1/2.
- As x increases, y increases at a slower rate compared to the line y = 2x.
Now, let's determine which variable (x or y) to integrate with respect to. Since the given curves are described explicitly as functions of y, it would be easier to integrate with respect to y.
To find the area of the region in question, follow these steps:
1. Determine the limits of integration for y.
- We need to find the y-values where the curves intersect.
- Set each pair of equations equal to each other and solve for x.
- For y = 7/x and y = 2x: 7/x = 2x
- Solve for x: x^2 = 7/2
- Taking the positive square root: x = √(7/2)
- We now have two limits of integration for y: 0 (from y = 0 to y = √(7/2)).
2. Set up the integration for the area:
- We know that y = 2x and y = x/2 intersect at y = 1.
- We need to determine the corresponding x-values for y = 1.
- For y = 2x: 1 = 2x (solve for x)
- x = 1/2
- For y = x/2: 1 = x/2 (solve for x)
- x = 2
- Now we have two limits of integration for x: 1/2 (from x = 1/2 to x = 2).
3. Integrate the curves with respect to y from y = 0 to y = √(7/2):
- Integrate y = 7/x with respect to y:
- ∫(1 to √(7/2)) (7/y) dy
- Integrate y = 2x with respect to y:
- ∫(√(7/2) to 1) 2x dy
- Integrate y = x/2 with respect to y:
- ∫(1 to 2) (x/2) dy
4. Evaluate the integrals and subtract the results to find the enclosed area.
Note: To find each individual integral, you would need to convert them to integrals with respect to x, since we solved for x in terms of y.