Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y = 4√x and y = 5 and 2y+2x = 6.

I would integrate along dy, since the boundary changes at x=1

That makes the integral easy:

∫[2,5] (y^2/4 - (3-y)) dy

If we integrate along dx, then we need
∫[-2,1] 5-(3-x) dx + ∫[1,25/4] 5-2√x dx

To sketch the region enclosed by the given curves and find its area, we can follow these steps:

1. Sketch the curves:
- The equation 2y = 4√x represents a parabola opening to the right with its vertex at the origin (0,0). To sketch the curve, you can plot some points by substituting different values of x and calculating the corresponding y values.
- The equation y = 5 represents a horizontal line at y = 5.
- The equation 2y + 2x = 6 can be rewritten as y = 3 - x. This represents a line with a slope of -1 and y-intercept of 3.

On a graph, you can plot these curves and label them accordingly.

2. Determine the intersection points:
To find the intersection points, we need to solve the equations simultaneously:
- Substituting y = 5 into the equation 2y = 4√x, we get 10 = 4√x. Simplifying, we get √x = 10/4 = 5/2. Squaring both sides, we have x = (5/2)² = 25/4.
- Substituting y = 5 into the equation y = 3 - x, we get 5 = 3 - x. Solving for x, we find x = -2.

So the intersection points are (-2, 5) and (25/4, 5).

3. Decide whether to integrate with respect to x or y:
To determine the correct integration bounds, we look at the x-values of the intersection points. The region is bounded between x = -2 and x = 25/4. Therefore, we will integrate with respect to x.

4. Set up the definite integral:
The area can be calculated as the integral of the top curve minus the bottom curve between the given bounds:
A = ∫[x=-2 to 25/4] (4√x - (3 - x)) dx.

5. Evaluate the integral:
By integrating the expression and evaluating it between the given bounds, we can find the area of the region.

Following these steps, you can sketch the region, find the intersection points, and then set up and evaluate the integral to determine the area of the enclosed region.