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=3sqrtx
and
y=3
and
2y+2x=5

its not 8

I would take vertical slices, that is, integrate with respect to x

But after looking at the sketch, I realize that we have to find the intersection of the line 2x+2y=5 and 2y = 3√x
equating 3√x = 5-2x and squaring both sides I got
4x^2 - 29x + 25=0
(x-1)(4x-25) =0
x = 1 or x = 25/4

also if y=3, then 6=3√x ----> x = 4

so we need ∫(3 - 5/2 + x dx from 0 to 1 + ∫(3 - 3x^(1/2) ) dx from 1 to 4

Can you finish it?

-4?? that is incorrect..hmm, what am i doing wrong?

To sketch the region enclosed by the given curves, let's first understand the shapes of the curves.

1. 2y = 3√x: This equation represents a curve in the coordinate plane. To visualize it, we can rewrite the equation as y = (3/2)√x. This means that when you plug in different values of x, you will get corresponding values of y. For example, when x = 4, y = (3/2)√4 = 3. The curve will gradually rise as x increases.

2. y = 3: This equation is a horizontal line parallel to the x-axis at y = 3. It does not change as x varies.

3. 2y + 2x = 5: This equation represents a straight line. We can rewrite it as y = (5 - 2x)/2. This means that as x changes, y will change linearly. The line will slope downwards as x increases.

Now, let's sketch the region enclosed by these curves:

1. The curve given by 2y = 3√x will start at the origin (0,0) and gradually rise as x increases.

2. The horizontal line y = 3 will be a straight line parallel to the x-axis, passing through y = 3.

3. The line given by 2y + 2x = 5 will intersect the y-axis at (0, 2.5) and the x-axis at (2.5, 0). It will slope downwards in the coordinate plane.

To determine the boundaries of the region, we need to find the points where these curves intersect.

1. To find the intersection point between 2y = 3√x and y = 3, we can substitute the value of y from the second equation into the first equation:
2(3) = 3√x
6 = 3√x
√x = 2
x = 4

So, the first curve intersects the line at x = 4.

2. To find the intersection point between y = 3 and 2y + 2x = 5, we can substitute the value of y from the first equation into the second equation:
2(3) + 2x = 5
6 + 2x = 5
2x = -1
x = -0.5

So, the second curve intersects the line at x = -0.5.

Now, we can see that the region is bounded by the curves y = 3 (the horizontal line), 2y = 3√x (the curve), and 2y + 2x = 5 (the line).

Since the curve and the line are defined as functions of y, we should integrate with respect to y to find the area of the region.

To find the area, we need to split the region into two parts: one part from x = -0.5 to 4, and another part from x = 4 to some upper limit that we still need to determine.

To find this upper limit, we can equate the line equation and the curve equation, 2y + 2x = 5 and 2y = 3√x, respectively:
3√x + 2x = 5
3√x = 5 - 2x
9x = 25 - 20x + 4x^2 [squaring both sides to remove the square root]
4x^2 - 29x + 25 = 0 [rearranged equation]

By solving this quadratic equation, we can find the value of x where the curve and line intersect at the right side of the region.

Once we have the upper limit of x, we can integrate the functions to find the area of each part and add them together to get the total area of the region.