find the area of the region enclosed by y= square root of x, y=x-2, and y=0.
i use f(x) - g(x) and got the answer to be 2, but my classmate got 8/3 do i'm really confuse don't know whats the answer.
Since y = x-2 starts below the x axis, I don't know whether to treat the region between the negative part of the parabola y = sqrtx and y = x-2 as the enclosed area. You get a much larger enclosed area if you only use y = +sqrtx to define the enclosed area. The end of the enclosed area is then where
x - 2 = sqrt x
x^2 -4x + 4 = x
(x-4)(x-1) = 0
x = 4 is the upper limit of x-integration if you don't use the negative part of the y = sqrtx curve.
To find the area of the region enclosed by the given curves, you need to find the points of intersection of the curves and then integrate the difference of the functions.
First, let's find the x-coordinates of the points where the curves intersect:
1. Set the two functions equal to each other:
√x = x - 2
2. Square both sides of the equation to eliminate the square root:
x = (x - 2)^2
3. Expand the right side:
x = x^2 - 4x + 4
4. Rearrange the equation:
0 = x^2 - 5x + 4
5. Factor the quadratic equation:
0 = (x - 1)(x - 4)
This gives us two critical x-values for intersection: x = 1 and x = 4.
Now, let's evaluate the area between the curves:
∫[1, 4] (f(x) - g(x)) dx
1. Calculate the difference between the two functions:
f(x) - g(x) = (√x) - (x - 2)
= √x - x + 2
2. Integrate the difference over the interval [1, 4]:
∫[1, 4] (√x - x + 2) dx
3. To evaluate this integral, we can split it into three separate integrals:
∫[1, 4] (√x) dx - ∫[1, 4] (x) dx + ∫[1, 4] (2) dx
4. Calculate each integral separately:
∫[1, 4] (√x) dx = (2/3)x^(3/2) | [1, 4]
= (2/3)(4^(3/2) - 1^(3/2))
= (2/3)(8 - 1)
= 14/3
∫[1, 4] (x) dx = (1/2)x^2 | [1, 4]
= (1/2)(4^2 - 1^2)
= (1/2)(16 - 1)
= 15/2
∫[1, 4] (2) dx = 2x | [1, 4]
= 2(4 - 1)
= 6
5. Calculate the total area:
Area = ∫[1, 4] (√x - x + 2) dx
= ∫[1, 4] (√x) dx - ∫[1, 4] (x) dx + ∫[1, 4] (2) dx
= 14/3 - 15/2 + 6
= 8/3
Therefore, the correct area of the region enclosed by the curves y = √x, y = x - 2, and y = 0 is 8/3 units^2.