find the area of the region:
y=4
y=2sqrt(x)
y=3-x
first two hit at (4,4)
curve 2 and 3 hit at
3 - x = 2 sqrt x
x + 2 sqrt x = 3
let z = sqrt x
z^2 + 2 z - 3 = 0
(z+3)(z-1) = 0
sqrt x = -3 or sqrt x = 1
so x = 1, y = 2
line 3 and line 1 hit at
(-1, 4)
draw all that on your sketch
now from x = -1 to x = 1
integrate [ 4 - 3+x) or in other words
(1+x) dx
and from x = 1 to x = 4
integrate [ 4 - 2 x^.5) dx
and add the two results
or, you can integrate over dy, and avoid the split:
In this case, the two lower bounding curves are
x = y^2/4
x = 3-y
∫[2,4] y^2/4 -(3-y) dy = 14/3
Checking with Damon's method gives
∫[-1,1] 4-(3-x) dx + ∫[1,4] 4-2√x dx
= 2 + 8/3
= 14/3
So, each method has its complications.
To find the area of the region enclosed by the given curves, we need to determine the points where these curves intersect and their respective bounding limits. Let's start by finding the points of intersection:
1. Set the equations equal to each other to find the x-values where the curves intersect:
4 = 2√(x) -> 2√(x) - 4 = 0
4 = 3 - x -> 3 - x - 4 = 0
2. Solve each equation separately:
For the first equation, let's simplify it:
2√(x) - 4 = 0
2√(x) = 4
√(x) = 2
Squaring both sides to eliminate the square root:
x = 4
For the second equation:
3 - x - 4 = 0
-x = 1
x = -1
So the curves intersect at x = 4 and x = -1.
3. Determine the y-values for each intersection point by substituting the x-values into one of the original equations.
For x = 4:
y = 2√(4) = 4
For x = -1:
y = 3 - (-1) = 4
Next, we find the bounds of the region for integration. The upper bound will be the curve with the highest y-values, which is y = 4, and the lower bound will be the curve with the lowest y-values, which is y = 2√(x).
Therefore, the area of the region can be found by integrating the difference between the two curves with respect to x from -1 to 4:
Area = ∫[from -1 to 4] (4 - 2√(x)) dx
Evaluating this integral will give us the area of the region.