The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work.

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1. Find the area of R.
2. The region R is the base of a solid. For each y, where 0<=y<=2, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 2y. Write, but do not evaluate, an integral expression that gives the volume of the solid.
3. There is a point P on the graph of f at which the line tangent to the graph of f is perpendicular to the graph of g. Find the coordinates of point P.

1. To find the area of the region R, we need to find the points of intersection of the graphs of f(x) and g(x), and then integrate the difference between them over that interval.

First, let's find the points of intersection:
f(x) = g(x)
√x = 6 - x

Squaring both sides:
x = (6 - x)^2
x = 36 - 12x + x^2

Rearranging the equation:
x^2 - 13x + 36 = 0

Factoring the quadratic equation:
(x - 4)(x - 9) = 0

So, x = 4 and x = 9 are the points of intersection.

Next, we will integrate the difference between f(x) and g(x) over the interval [4, 9]:
∫[4,9] (f(x) - g(x)) dx

∫[4,9] (√x - (6 - x)) dx
= ∫[4,9] (√x - 6 + x) dx
= ∫[4,9] (√x + x - 6) dx

The integral can be split into three separate integrals:
∫[4,9] √x dx + ∫[4,9] x dx - ∫[4,9] 6 dx

Integrating each term separately:
∫[4,9] x^(1/2) dx = (2/3)x^(3/2) | [4,9] = (2/3)(9^(3/2) - 4^(3/2))
∫[4,9] x dx = (1/2)x^2 | [4,9] = (1/2)(9^2 - 4^2)
∫[4,9] 6 dx = 6x | [4,9] = 6(9 - 4)

Putting it all together:
Area of R = (2/3)(9^(3/2) - 4^(3/2)) + (1/2)(9^2 - 4^2) + 6(9 - 4)

2. To find the volume of the solid whose cross-section is a rectangle of base in R and height 2y, we need to integrate the area of each cross-section over the given range of y.

Let's call the integral expression for the volume V(y):
V(y) = ∫[0,2] A(y) dy

We know that the base of the rectangle lies in the region R, and the base width is given by the difference between the x-values of the points of intersection, which is (9 - 4) = 5. Therefore, the area of each cross-section is 5 * 2y = 10y.

Substituting this into the integral expression:
V(y) = ∫[0,2] 10y dy

3. To find the point P on the graph of f(x) where the tangent line is perpendicular to the graph of g(x), we need to find a point where the slopes of the tangent line to f(x) and the slope of g(x) are negative reciprocals of each other.

The slope of the tangent line to f(x) can be found by taking the derivative of f(x):
f'(x) = 1/(2√x)

The slope of g(x) is -1 (since it is a horizontal line).

So, we want to find a point where:
1/(2√x) = -1

Solving for x:
2√x = -1
√x = -1/2
x = (-1/2)^2
x = 1/4

To find the y-coordinate of the point P, we substitute x = 1/4 into f(x):
f(1/4) = √(1/4)
= 1/2

Therefore, the coordinates of point P are (1/4, 1/2).