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=5x^(1/2) , y=4 and 2y+1x=6

I've been trying this problem for about 3 hours.
Please help!!!!!

The region of interest is a triangle-ish area. The intersections are

(-2,4) (64/25,4) (1,5/2)

wrt x, divide region into two parts:

∫[-2,1] 4 - (6-x)/2 dx
+ ∫[1,64/25] 4 - 5/2 √x dx
= 9/4 + 27/25 = 333/100

wrt y,

∫[5/2,4] 4/25 y^2 - (6-2y) dy
= 333/100

Thank you so much!!!!!

no sweat. Hang in there and after a few hundred more of these, they'll be a piece of cake!

To sketch the region enclosed by the given curves and determine whether to integrate with respect to x or y, let's analyze the equations one by one:

1. Equation: 2y = 5x^(1/2)
This equation represents a curve in the form of a square root function. To determine whether to integrate with respect to x or y, we need to observe which variable is easier to isolate. In this case, it is easier to solve for x in terms of y:
x = (2y/5)^(2)
Thus, it is more convenient to integrate with respect to y.

2. Equation: y = 4
This equation represents a horizontal line at y = 4. It does not depend on x, so it does not affect the choice on which variable to integrate with. However, we include it for completeness in determining the limits of integration.

3. Equation: 2y + x = 6
This equation represents a line in slope-intercept form. We rewrite it as follows:
x = 6 - 2y
Since the variable x depends on y, it is easier to integrate with respect to y in this case as well.

To find the area of the region between these curves, we integrate with respect to y. The limits of integration are determined by the common y-values of the curves.

First, calculate the y-values at the points of intersection:
For the first two curves (Equations 1 and 2):
2y = 5x^(1/2)
4 = 5x^(1/2)
x^(1/2) = 4/5
x = (4/5)^(2) = 16/25

For the last two curves (Equations 2 and 3):
2y + x = 6
2(4) + x = 6
x = 6 - 8 = -2

Therefore, the limits of integration for y are from y = 0 (x-axis) to y = 4 (equation y = 4).

To calculate the area using integration, we integrate the "top" curve minus the "bottom" curve with respect to y:
A = ∫ [(6 - 2y) - (2y/5)^(2)] dy

Evaluating this integral will give you the area between the curves.

Please note that if at any stage you need help with the integration process, feel free to ask for assistance.