Sketch the region enclosed by 2y=5sqrtx, y=5 and 2y+3x=8.
Decide whether to integrate with respect to x or y, and then find the area of the region.
Where do you get stuck?
The curve and the sloping line intersect at (1,5/2).
I'd integrate on y, from 5/2 to 5
5/2 sqrt(x) - (8-3x)/2
Oops. Integrating on y we need x as f(y)
4y^2/25 - (8-2y)/3
To sketch the region enclosed by the given equations, let's first solve each equation for y:
1) Equation 1: 2y = 5√x
Dividing both sides by 2, we get:
y = (5/2)√x
2) Equation 2: y = 5
Now let's plot the graph of each equation:
For Equation 1, we know that as x increases, y increases. Since the square root function is always positive, y will be positive as well. This means that Equation 1 represents the upper half of the curve.
For Equation 2, y is a constant 5. Therefore, it represents a horizontal line at y = 5.
Next, let's analyze the third equation:
3) Equation 3: 2y + 3x = 8
Solving for y, we have:
2y = 8 - 3x
Dividing both sides by 2, we get:
y = (8 - 3x)/2
Simplifying further, we have:
y = 4 - (3/2)x
Now we have all the information needed to sketch the region:
- The curve represented by Equation 1 (y = (5/2)√x), which is the upper half of the curve.
- The horizontal line represented by Equation 2 (y = 5).
- The line represented by Equation 3 (y = 4 - (3/2)x).
Now, we need to find the points where these equations intersect:
- To find the intersection of Equations 1 and 2, we set them equal to each other:
(5/2)√x = 5
√x = 2
x = 4
Substituting x into Equation 2, we get: y = 5
- To find the intersection of Equations 2 and 3, we set them equal to each other:
5 = 4 - (3/2)x
(3/2)x = -1
x = -2/3
Substituting x into Equation 2, we get: y = 5
Now let's plot these points on a graph:
- Point A: (4, 5)
- Point B: (-2/3, 5)
Now, we can sketch the region enclosed by these curves:
C (4, 5)
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B (-2/3, 5)
------------------------x-axis---------------
The region enclosed by these curves is the area between the curve represented by Equation 1, the horizontal line represented by Equation 2, and the line represented by Equation 3. To find the area of this region, we will integrate with respect to x.
To determine the limits of integration, we need to find the x-values where these curves intersect. We already found these values to be 4 and -2/3.
Therefore, we will integrate with respect to x from -2/3 to 4.
The area of the region can be calculated as follows:
Area = ∫[from -2/3 to 4] [(5/2)√x - (4 - (3/2)x)] dx
We can now integrate this expression to find the area of the region enclosed by these curves.
To sketch the region enclosed by the given equations, we first need to determine the intersection points of the three curves. Let's break down the process step by step:
1. Find the intersection points of the curves 2y = 5√x and y = 5 by equating the two equations:
2y = 5√x (Equation 1)
y = 5 (Equation 2)
By substituting Equation 2 into Equation 1, we can solve for x:
2(5) = 5√x
10 = 5√x
Dividing both sides by 5 yields:
2 = √x
Squaring both sides of the equation to eliminate the square root:
4 = x
So, the intersection point for these two curves is (4, 5).
2. Now let's find the intersection points of the curves y = 5 and 2y + 3x = 8. We can substitute y = 5 into the second equation:
2(5) + 3x = 8
10 + 3x = 8
3x = -2
x = -2/3
Thus, the intersection point for these two curves is (-2/3, 5).
3. Now, we can plot these points and sketch the region enclosed by these curves. The three curves are:
- Curve 1: y = 5
- Curve 2: 2y = 5√x
- Curve 3: 2y + 3x = 8
The point obtained from the first step, (4, 5), lies on the curve 2. The point obtained from the second step, (-2/3, 5), lies on the curve 3.
Therefore, curve 1 lies above the region, curve 2 lies to the right of the region, and curve 3 forms the boundary of the region.
The sketch will look like this:
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| Boundary (Curve 3)
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| Region
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|...................
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| Curve 1 (y = 5)
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Now, let's find the area of this region.
The region is bounded from left to right by the x-values -2/3 and 4, and from bottom to top by the y-value 5. Since the region is vertical, we will integrate with respect to x to find the area.
To do this, we will need to integrate the equation of the curve that forms the top boundary of the region, and subtract the integral of the equation of the curve that forms the bottom boundary of the region.
The integral for the curve 2 can be found as follows:
2y = 5√x
y = (5√x)/2
Therefore, the integral for the top boundary is:
∫ [(5√x)/2] dx
The integral for the bottom boundary (curve 1) is simply the constant y = 5.
Now, we can calculate the area of the region by subtracting these two integrals:
Area = ∫ [(5√x)/2] dx - ∫ 5 dx
Area = (1/2) ∫ 5√x dx - 5 ∫ dx
Area = (1/2) [5(2/3)x^(3/2)] - 5x + C
Area = 5/3 x^(3/2) - 5x + C
To find the definite integral between the limits of x = -2/3 and x = 4, we can substitute these values into the equation and subtract the result:
Area = [(5/3)(4)^(3/2) - 5(4)] - [(5/3)(-2/3)^(3/2) - 5(-2/3)]
Area = [(5/3)(8) - 20] - [(5/3)(-4/3) - 10/3]
Area = (40/3 - 20) - (-20/9 - 10/3)
Area = 40/3 - 60/3 + 20/9 + 10/3
Area = 130/9
Therefore, the area of the region enclosed by the curves is 130/9 square units.