Find the area in the first quadrant bounded by the curve y=(9-x)^1/2 and the x- and y-axis.
To find the area in the first quadrant bounded by the given curve, we will integrate the function with respect to x.
1. Start by noting the limits of the integral. Since we are considering the first quadrant, the x-value ranges from 0 to the x-coordinate where the curve intersects the x-axis.
2. To determine the x-coordinate where the curve intersects the x-axis, we set y = 0 in the equation y = (9 - x)^(1/2) and solve for x:
(9 - x)^(1/2) = 0
Squaring both sides, we get:
9 - x = 0
x = 9
3. Now we can set up the integral to find the area A:
A = ∫[0 to 9] (9 - x)^(1/2) dx
4. Using the power rule for integration, we can rewrite the integral as:
A = ∫[0 to 9] (9 - x)^(1/2) dx = [2/3(9 - x)^(3/2)] from 0 to 9
5. Evaluate the integral:
A = (2/3)(9 - 9)^(3/2) - (2/3)(9 - 0)^(3/2)
= (2/3)(0) - (2/3)(9)^(3/2)
= -2/3 * 27^(3/2)
= -2/3 * 27 * (3)^(1/2)
6. Since we are dealing with area, the result cannot be negative. Therefore, we take the absolute value of the expression:
A = |(-2/3 * 27 * (3)^(1/2))|
= 2/3 * 27 * (3)^(1/2)
= 54 * (3)^(1/2)
Therefore, the area in the first quadrant bounded by the curve y = (9 - x)^(1/2) and the x- and y-axis is 54 * (3)^(1/2) square units.