Decide whether to integrate with respect to x or y. Then find the area of the region: 2y=5sqrt(x), y=4, and 2y+2x=7.
To determine whether to integrate with respect to x or y, we need to analyze the given equations and determine the bounds for integration.
Let's start by graphing the equations to visualize the region.
1. Equation: 2y = 5√(x)
Solving for y: y = 5√(x)/2
This equation represents a curve in the first quadrant, as it is always positive.
2. Equation: y = 4
This equation represents a horizontal line parallel to the x-axis at y = 4.
3. Equation: 2y + 2x = 7
Solving for y: y = (7-2x)/2
This equation represents a line with a negative slope.
To find the area of the region, we need to determine the bounds for integration in both x and y.
Considering the graph and equations:
- The region is bound by the curve y = 5√(x)/2 and the line y = 4 on the top.
- The region is bound by the curve y = 5√(x)/2 and the line y = (7-2x)/2 on the bottom.
Now, we can decide whether to integrate with respect to x or y.
If we integrate with respect to y, we need to find the bounds for y in terms of x. From the equations, we can see that the upper bound for y is given by y = 4, and the lower bound is given by y = (7-2x)/2. Thus, the integration bounds in terms of y are 4 and (7-2x)/2.
If we integrate with respect to x, we need to find the bounds for x in terms of y. From the equations, we can see that the lower bound for x is given by the curve y = 5√(x)/2, and the upper bound is given by the line y = (7-2x)/2. Thus, the integration bounds in terms of x are [(2y/5)^2] and (7-2y)/2.
To choose whether to integrate with respect to x or y, we should evaluate which set of bounds are easier to work with. In this case, it seems that integrating with respect to x would be simpler, as the bounds only involve linear equations.
Thus, we will integrate with respect to x.
Now, let's determine the area by finding the definite integral:
A = ∫[x̂₁, x̂₂] ([(7-2y)/2] - [(2y/5)^2]) dx,
where x̂₁ and x̂₂ are the bounds in terms of x.
We can evaluate this integral to find the area of the region.