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

I'd use horizontal strips to avoid having to switch curves. That makes the area

∫[5/2,5] (4/25 y^2 - (8-2y)/3) dy

Of course, you could use vertical strips, but it's more work
∫[0,1] (5 - (8-3x)/2) dx + ∫[1,4] (5 - 5/2 √x) dx

check to make sure I got the intersections right.

To sketch the region enclosed by the three curves and find its area, let's start by graphing each of the equations separately.

1. Equation: 2y = 5√x
To get a clearer picture, let's square both sides of the equation to eliminate the square root:

(2y)^2 = (5√x)^2
4y^2 = 25x
y^2 = (25/4)x

This equation represents a parabola that opens to the right, with its vertex at the origin (0,0).

2. Equation: y = 5
This equation represents a horizontal line parallel to the x-axis, passing through y = 5.

3. Equation: 2y + 3x = 8
Rearranging the equation, we have:
2y = 8 - 3x
y = 4 - (3/2)x

Now, let's graph these equations and shade the region that is enclosed by these curves:

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Based on the graph, it is clear that the region is enclosed between the parabola (y^2 = (25/4)x) and the horizontal line (y = 5). The right side of the region is defined by the line (y = 4 - (3/2)x).

To find the area of this region, we need to integrate with respect to x.

To determine the limits of integration, we find the x-values where the curves intersect:

1. y^2 = (25/4)x and y = 5
Setting these equations equal, we get:
(25/4)x = 25

Simplifying, we find:
x = 4

2. y^2 = (25/4)x and 4 - (3/2)x = 5
Setting these equations equal, we get:
(25/4)x = 9

Simplifying, we find:
x = 36/25

Therefore, the limits of integration will be x = 36/25 to x = 4.

Now, let's integrate the difference between the two curves with respect to x to find the area:

Area = ∫[(4 - (3/2)x) - 5] dx
= ∫(4 - (3/2)x - 5) dx
= ∫(4 - (3/2)x - 5) dx
= ∫(- (3/2)x - 1) dx
= [- (3/4)x^2 - x] evaluated from x = 36/25 to x = 4

Evaluating this integral, we find:
Area = [- (3/4)(4)^2 - (4)] - [- (3/4)(36/25)^2 - (36/25)]

Simplifying further, we get:
Area = [- (3/4)(16) - (4)] - [- (3/4)(1296/625) - (36/25)]
Area = (-12 -4) - (- 3888/100 - 72/25)
Area = -16 + (3888/100 + 72/25)

Calculating further, we have:
Area = -16 + (38880 + 288)/500
Area = -16 + 39168/500
Area = -16 + 78.336
Area = 62.336

Hence, the area of the region enclosed by the given curves is approximately 62.336 square units.

To sketch the region enclosed by the given equations, we need to find the points where they intersect. Let's start by solving for the points of intersection:

1. Equating the first equation, 2y = 5√x, to the second equation, y = 5:
2(5) = 5√x
10 = 5√x
√x = 2
x = 4

2. Substituting the value of x in the third equation, 2y + 3x = 8:
2y + 3(4) = 8
2y + 12 = 8
2y = -4
y = -2

So, the coordinates of the point of intersection are (4, -2).

Now, let's determine whether to integrate with respect to x or y. Looking at the given equations, the region is bounded by the functions y = 5 and 2y = 5√x. Notice that y is a better variable to integrate with since both functions are expressed in terms of y.

To find the area of the region, we'll integrate with respect to y.

The bounds of integration will be the y-values that enclose the region, which are y = -2 and y = 5.

Now, let's set up the integral to find the area:

Area = ∫[from y = -2 to y = 5] [upper curve - lower curve] dy

The upper curve is given as y = 5 and the lower curve is given as 2y = 5√x. As x is not fixed, we need to express the lower curve equation in terms of y and x to solve the integral properly.

Solving 2y = 5√x for x:
2y/5 = √x
(x = (2y/5)^2 = (4y^2)/25)

The integral is:
Area = ∫[from y = -2 to y = 5] [5 - (4y^2)/25] dy

Evaluating this integral will give you the area of the region enclosed by the given equations.