Question:

Consider the function f(x) = 2x + 1 defined for x ≥ 0. The graph of f is shown below:

(a) Find the area of the region bounded by the graph of f(x) and the x-axis, in the interval [1, 3].

(b) Find the x-coordinate(s) of the point(s) on the graph of f(x) where the tangent line is parallel to the x-axis.

(c) Find the value of k such that the area of the region bounded by the graph of f(x) and the x-axis, in the interval [0, k], is equal to 9.

Answer:

(a) To find the area of the region bounded by the graph of f(x) and the x-axis in the interval [1, 3], we need to find the definite integral of f(x) in this interval.

The definite integral is given by:

∫[a, b] f(x) dx

In this case, a = 1 and b = 3. Substituting these values into the definite integral:

∫[1, 3] f(x) dx = ∫[1, 3] (2x + 1) dx

To evaluate this integral, we will use the power rule of integration:

∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1

Applying the power rule, we have:

∫[1, 3] (2x + 1) dx = [(2x^2)/2 + x] | [1,3] = (2(3)^2)/2 + 3 - (2(1)^2)/2 - 1

Simplifying, we get:

Area = (18/2 + 3) - (2/2 + 1) = 12 + 3 - 1 = 14

Therefore, the area of the region bounded by the graph of f(x) and the x-axis in the interval [1, 3] is 14 square units.

(b) To find the x-coordinate(s) of the point(s) on the graph of f(x) where the tangent line is parallel to the x-axis, we need to find the value(s) of x where f'(x) = 0.

First, let's find f'(x) by differentiating f(x):

f'(x) = d/dx (2x + 1) = 2

Since f'(x) is a constant (2), it is never equal to 0. Therefore, there are no x-coordinate(s) where the tangent line to the graph of f(x) is parallel to the x-axis.

(c) To find the value of k such that the area of the region bounded by the graph of f(x) and the x-axis in the interval [0, k] is equal to 9, we need to solve the following equation:

∫[0, k] (2x + 1) dx = 9

Using the definite integral formula and the power rule of integration, we have:

[(2x^2)/2 + x] | [0, k] = (2k^2)/2 + k - (2(0)^2)/2 - 0 = k^2 + k - 0 = k^2 + k

Simplifying, we have:

k^2 + k = 9

Rearranging the equation, we get:

k^2 + k - 9 = 0

This is a quadratic equation that can be factored:

(k - 2)(k + 3) = 0

Setting each factor equal to 0, we find two possible values for k:

k - 2 = 0 => k = 2 k + 3 = 0 => k = -3

Since k represents a length and it cannot be negative, we discard the solution k = -3.

Therefore, the value of k such that the area of the region bounded by the graph of f(x) and the x-axis in the interval [0, k] is equal to 9 is k = 2.