Answer:
G. 8/3
Explanation:
To find the area of the region in the first quadrant bounded by the curves √x, y=2 and the y-axis, we need to integrate the function that gives the height of the region at each point along the x-axis. The height of the region is given by the difference between the curve y=2 and the curve y=√x.
We can write the integral for the area A as follows:
A = ∫[0,4] (2-√x) dx
The limits of integration are from 0 to 4 because the curves intersect at x=4. We integrate with respect to x because the curves are functions of x.
To evaluate the integral, we first use the power rule of integration to simplify the expression inside the integral:
A = ∫[0,4] (2-√x) dx = [2x - (2/3)x^(3/2)]|[0,4]
Now, we substitute the limits of integration:
A = [2(4) - (2/3)(4)^(3/2)] - [2(0) - (2/3)(0)^(3/2)]
A = 8 - (16/3)
A = 8/3
Therefore, the area of the region in the first quadrant bounded by the curves √x, y=2 and the y-axis is 8/3 square units.