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=5, and 2y+3x=6
I tried this many times and i am getting it wrong. Please show me how you got to the answer.
To sketch the region enclosed by the given curves, we need to understand the equations and their corresponding graphs. Let's go step by step:
1. First, let's examine the equation 2y = 3√(x). To make things simpler, we can rewrite it as y = (3/2)√(x).
- If we plot this equation on a graph, we will see a curve starting from the origin and opening towards the positive x-axis.
- The curve represents all the points where the equation is satisfied. However, note that y values can only be positive or zero since we have a square root.
2. Next, let's consider the equation y = 5. This equation is a horizontal line passing through y = 5 on the y-axis.
- This line represents all the points where the equation is satisfied, i.e., y is always equal to 5.
3. Finally, let's examine the equation 2y + 3x = 6. We can rearrange it as y = (6 - 3x)/2, which simplifies to y = 3 - (3/2)x.
- If we plot this equation on a graph, we will see a straight line with a negative slope and y-intercept at y = 3.
- This line represents all the points where the equation is satisfied.
Now, let's determine the region enclosed by these curves:
- The curve y = (3/2)√(x) starts from the origin and opens towards the positive x-axis.
- The line y = 5 is a horizontal line.
- The line y = 3 - (3/2)x is a straight line with a negative slope.
Considering the range of x values for which these curves exist, we can see that the region is enclosed between the points where these curves intersect. Let's find those points:
1. Setting the equation (3/2)√(x) = 5, we can solve for x:
(3/2)√(x) = 5
√(x) = 10/3 (by multiplying both sides by 2/3)
x = (10/3)^2 = 100/9
2. To find the x-coordinate where the line y = 3 - (3/2)x intersects the curve y = (3/2)√(x), we can substitute y into the equation:
y = 3 - (3/2)x
(3/2)√(x) = 3 - (3/2)x
Multiply both sides by 2:
3√(x) = 6 - 3x
√(x) = 2 - x
Square both sides:
x = 4 - 4x + x^2
Rearrange the equation:
x^2 - 5x + 4 = 0
Using factoring or the quadratic formula, we can find that the solutions to this equation are x = 1 and x = 4.
Now, we know the x-coordinates of the points of intersection: x = 1 and x = 4. To find the corresponding y-values, we substitute these x-values into the equations:
For x = 1:
- y = (3/2)√(1) = 3/2
- y = 3 - (3/2)(1) = 3 - 3/2 = 3/2
For x = 4:
- y = (3/2)√(4) = 3
- y = 3 - (3/2)(4) = 3 - 6 = -3
Since the region is enclosed between the x-values 1 and 4, and the corresponding y-values, we need to integrate with respect to x.
To find the area of the region, we need to integrate the area between the curves from x = 1 to x = 4.
The area can be calculated using the integral: ∫[1 to 4] [(3 - (3/2)x) - (3/2)√(x)] dx.
Evaluating this integral will give us the area of the enclosed region.