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=3sqrt(x) , y=3 and 2y+3x=6.
To sketch the region enclosed by the given curves, we first need to find the points of intersection between the curves.
1. Curve 1: 2y=3√(x)
To find the points of intersection of this curve, let's set y=3 and solve for x:
2(3) = 3√(x)
6 = 3√(x)
2 = √(x)
x = 4
So, this curve intersects the line y=3 at x=4.
2. Curve 2: 2y+3x=6
Again, let's set y=3 and solve for x:
2(3) + 3x = 6
6 + 3x = 6
3x = 0
x = 0
This curve intersects the line y=3 at x=0.
Now, let's plot these points of intersection on a graph:
- The curve 2y=3√(x) intersects the line y=3 at x=4.
- The line 2y+3x=6 intersects the line y=3 at x=0.
Next, we determine the boundaries of the region. Looking at the sketch, we can see that the graph of 2y=3√(x) is above the line 2y+3x=6 in the region of interest. Therefore, the area is enclosed between the curves 2y=3√(x) and y=3, bounded by x=0 and x=4.
To find the area of this region, we need to integrate with respect to x (from x=0 to x=4):
Area = ∫(from 0 to 4) (upper curve - lower curve) dx
The upper curve is y = 3 and the lower curve is y = 2√(x)/3.
So, the area is given by:
Area = ∫(from 0 to 4) (3 - 2√(x)/3) dx
To evaluate this integral and find the area, we can use the fundamental theorem of calculus or any other suitable integration method.