Sketch the region enclosed by 2𝑦=5𝑥‾‾√,𝑦=5, and 2𝑦+2𝑥=7. Decide whether to integrate with respect to 𝑥 or 𝑦, and then find the area of the region.
Cannot sketch on these posts.
To sketch the region enclosed by the given equations, let's first examine each equation separately:
1. 2𝑦 = 5𝑥‾‾√
We can rewrite this equation as 𝑦 = (5𝑥‾‾√)/2. This equation represents a curve that starts from the origin (0,0) and increases as x increases.
2. 𝑦 = 5
This equation represents a horizontal line at 𝑦 = 5, which is a straight line parallel to the x-axis.
3. 2𝑦 + 2𝑥 = 7
This equation can be rewritten as 𝑦 = (7 - 2𝑥)/2. This equation represents a line with a negative slope.
Now, let's plot these equations on a graph:
1. Plot the equation 𝑦 = (5𝑥‾‾√)/2. Choose some x-values, substitute them into the equation to find the corresponding y-values, and plot the points. Make sure to include the point (0,0) as well.
2. Draw the line 𝑦 = 5. This line is a horizontal line passing through 𝑦 = 5.
3. Plot the line 𝑦 = (7 - 2𝑥)/2. Choose some x-values, substitute them into the equation to find the corresponding y-values, and plot the points.
Now, we can see that the region enclosed by these curves looks like a triangle.
To determine whether we should integrate with respect to 𝑥 or 𝑦, we need to identify the bounds of integration. We can do this by finding the points where the curves intersect.
1. Set 𝑦 = (5𝑥‾‾√)/2 equal to 𝑦 = 5 and solve for 𝑥:
(5𝑥‾‾√)/2 = 5
5𝑥‾‾√ = 10
𝑥‾‾√ = 2
𝑥 = 4
2. Set 𝑦 = (7 - 2𝑥)/2 equal to 𝑦 = 5 and solve for 𝑥:
(7 - 2𝑥)/2 = 5
7 - 2𝑥 = 10
-2𝑥 = 3
𝑥 = -3/2
From our graph, we can see that the curves intersect at 𝑥 = -3/2 and 𝑥 = 4.
Since the region is bounded by the curves 2𝑦 = 5𝑥‾‾√, 𝑦 = 5, and 2𝑦 + 2𝑥 = 7, we should integrate with respect to 𝑦 in this case.
Now, to find the area of the region, we need to find the integral of the difference between the upper and lower curves with respect to 𝑦:
∫[𝑦=5 to 𝑦=0] [(5𝑥‾‾√)/2 - (7 - 2𝑥)/2] 𝑑𝑦
Simplifying this integral will give us the area of the region enclosed by the given equations.