AP Calculus AB Exam Question:

Consider the functions f(x) = x^2 - 3 and g(x) = 2x - 1 on the interval [0, 2]. Find the area of the region bounded by the curves f(x) and g(x).

Solution:

To find the area between two curves, we need to determine the points of intersection between the curves and then integrate the difference between the curves over the interval.

Step 1: Finding the points of intersection:

To find the points of intersection, we set the two equations equal to each other:

x^2 - 3 = 2x - 1

Rearranging the equation, we get:

x^2 - 2x - 2 = 0

Factoring the quadratic equation, we have:

(x - 2)(x + 1) = 0

Setting each factor equal to zero, we get:

x - 2 = 0 --> x = 2 x + 1 = 0 --> x = -1

So the two curves intersect at x = -1 and x = 2.

Step 2: Determine the upper and lower curves:

To find the upper and lower curves on the interval [0, 2], we evaluate each function at the endpoints and the points of intersection:

At x = 0: f(0) = 0^2 - 3 = -3 g(0) = 2(0) - 1 = -1

At x = -1: f(-1) = (-1)^2 - 3 = -2 g(-1) = 2(-1) - 1 = -3

At x = 2: f(2) = 2^2 - 3 = 1 g(2) = 2(2) - 1 = 3

From the evaluations, we see that f(x) is the upper curve on the interval [0, 2], and g(x) is the lower curve.

Step 3: Set up and evaluate the integral:

The area between two curves can be found by integrating the difference between the upper and lower curves. In this case, we integrate the difference between f(x) and g(x) over the interval [0, 2]:

∫[0,2] (f(x) - g(x)) dx

∫[0,2] ((x^2 - 3) - (2x - 1)) dx

Simplifying,

∫[0,2] (x^2 - 2x - 2) dx

To evaluate the integral, we use the power rule:

∫ x^n dx = (x^(n+1))/(n+1)

Using the power rule, we have:

∫[0,2] (x^2 - 2x - 2) dx = (x^3/3 - x^2 - 2x) | [0,2]

Evaluating the antiderivative at the upper and lower limits:

= ((2)^3/3 - (2)^2 - 2(2)) - ((0)^3/3 - (0)^2 - 2(0))

= (8/3 - 4 - 4) - (0/3 - 0 - 0)

= (8/3 - 8) - (0 - 0 - 0)

= -8/3

Therefore, the area of the region bounded by the curves f(x) and g(x) on the interval [0, 2] is -8/3 square units.