I'm having trouble on this question:
Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.
Sketch the region. It has vertices at (0,0), (2,0), (4,2)
You can get the area by using vertical or horizontal strips. Using vertical strips, they all lie below the curve √x, but the lower boundary changes from y=0 to y=x-2 at x=2. So, the area is
∫[0,2] √x dx + ∫[2,4] √x - (x-2) dx
= 4√2/3 + 10/3 - 4/3 √2 = 10/3
Or, you could just subtract the area of the triangle from the whole area under the curve:
∫[0,4] √x dx - (1/2)(2*2)
= 16/3 - 2 = 10/3
Using horizontal strips, the left boundary is the parabola x=y^2, and the right boundary is x = y+2, so the area is
∫[0,2] (y+2)-y^2 dy = 10/3
To find the area of the region bounded by the curve, the x-axis, and the line, you need to set up an integral.
First, we need to determine the x-values where the curve y = sqrt(x) and the line y = x - 2 intersect. Setting the two equations equal to each other:
sqrt(x) = x - 2
Squaring both sides of the equation:
x = (x - 2)^2
Expanding and rearranging:
x = x^2 - 4x + 4
Bringing all terms to one side:
x^2 - 5x + 4 = 0
Factoring the quadratic equation:
(x - 4)(x - 1) = 0
So the two x-values where the curve and the line intersect are x = 4 and x = 1.
To find the area, we need to integrate the difference between the two curves with respect to x from x = 1 to x = 4.
The curve y = sqrt(x) is the upper curve, and the line y = x - 2 is the lower curve. Thus, the integral can be set up as:
A = ∫[1 to 4] (sqrt(x) - (x - 2)) dx
Using the power rule of integration, we can solve the integral:
A = ∫[1 to 4] (x^(1/2) - x + 2) dx
Evaluating the integral:
A = [2/3 * x^(3/2) - 1/2 * x^2 + 2x] [1 to 4]
Plugging in the limits of integration:
A = [2/3 * (4^(3/2)) - 1/2 * (4^2) + 2(4)] - [2/3 * (1^(3/2)) - 1/2 * (1^2) + 2(1)]
Simplifying:
A = [2/3 * 8 - 1/2 * 16 + 8] - [2/3 * 1 - 1/2 * 1 + 2]
A = [16/3 - 8 + 8] - [2/3 - 1/2 + 2]
A = [16/3] - [1/6]
A = (96 - 3)/18
A = 93/18
Simplifying, the area of the region is approximately 5.17 square units.