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=3(x^(1/2)) , y=5 and 2y+3x=6
The graphs intersect at (1,3/2), (100/9,5), (-4/3,5)
You can integrate over either x or y. Since one boundary is a horizontal line, integrating over x will require breaking the area into two pieces:
a = ∫[-4/3,1] 5-(6-3x)/2 dx + ∫[1,100/9] 5- 3/2 √x dx
= 49/12 + 392/27 = 2009/108
integrating over y keeps to a single region, but the widths are the difference between the two curves:
a = ∫[3/2,5] 4y^2/9 - (6-2y)/3 dy
= 2009/108
To sketch the region enclosed by the curves, we need to find the intersection points of the curves first.
First curve: 2y = 3√(x)
Rearranging the equation, we get: y = (3/2)√(x)
Second curve: y = 5 (a horizontal line)
Third curve: 2y + 3x = 6
Rearranging the equation, we get: y = (6-3x)/2
To find the intersection points, we need to set the equations for the curves equal to each other and solve for the variables.
(3/2)√(x) = 5
Simplifying, we get: √(x) = 10/3
Squaring both sides, we get: x = 100/9
(6-3x)/2 = 5
Simplifying, we get: 6 - 3x = 10
Solving for x, we get: x = -4/3
Since x cannot be negative for this problem, we discard the negative value and consider x = -4/3.
Now, let's sketch the region enclosed by the curves on a coordinate plane.
The graph of y = (3/2)√(x) is a curve that starts from the origin and gradually increases to the right.
The graph of y = 5 is a horizontal line at y = 5.
The graph of y = (6-3x)/2 is a straight line with a y-intercept at (0,3) and a slope of -3/2.
The region enclosed by these curves is a triangle bounded by the x-axis, the curve y = (3/2)√(x), and the line y = (6-3x)/2.
To find the area of the region, we need to determine the limits of integration. Since the region is bounded by the curves vertically, we should integrate with respect to x.
Thus, we need to find the limits of integration.
1. The leftmost curve is y = (3/2)√(x). We need to find the x-coordinate of the point where it intersects the x-axis.
(3/2)√(x) = 0
x = 0
2. The rightmost curve is y = (6-3x)/2. We need to find the x-coordinate of the point where it intersects the x-axis.
(6-3x)/2 = 0
6-3x = 0
x = 2
Therefore, the limits of integration are x = 0 to x = 2.
Now, we can calculate the area of the region using the definite integral.
Area = ∫[0 to 2] [(6-3x)/2 - (3/2)√(x)] dx
By evaluating this integral, we can find the area of the region enclosed by the given curves.