Question:

Consider the functions f(x) = x^2 and g(x) = 2x - 1.

(a) Find the points of intersection of the graphs of f(x) and g(x).

(b) Find the area of the region bounded by the graphs of f(x) and g(x) over the interval [0, 2].

Answer:

(a) To find the points of intersection, we need to set the two functions equal to each other and solve for x:

x^2 = 2x - 1

Rearranging the equation, we have:

x^2 - 2x + 1 = 0

Using factoring, we can rewrite the equation as:

(x - 1)(x - 1) = 0

This implies that x = 1 is the point of intersection of the two graphs.

(b) To find the area of the region bounded by the graphs of f(x) and g(x) over the interval [0, 2], we need to find the definite integral of the difference between the two functions over this interval:

Area = ∫[0, 2] (f(x) - g(x)) dx

= ∫[0, 2] (x^2 - (2x - 1)) dx

= ∫[0, 2] (x^2 - 2x + 1) dx

To integrate, we use the power rule:

∫ x^n dx = (1/(n+1)) * x^(n+1) + C

Applying the power rule to our integral:

= (1/3) * x^3 - x^2 + x | [0, 2]

Evaluating the integral and subtracting the lower limit from the upper limit:

= [(1/3) * (2^3) - 2^2 + 2] - [(1/3) * (0^3) - 0^2 + 0]

= [(8/3) - 4 + 2] - [0 - 0 + 0]

= (8/3) - 4 + 2

= 2/3

Therefore, the area of the region bounded by the graphs of f(x) and g(x) over the interval [0, 2] is 2/3 square units.