AP Calculus AB Exam Question:

Let $f(x) = 2x^3 - 3x^2 + 4$ be a function defined on the closed interval $[0, 2]$. The graph of $f(x)$ is shown below:

a) Find the average value of $f(x)$ on the interval $[0, 2]$.

b) Determine the x-coordinate(s) at which the average value of $f(x)$ on the interval $[0, x]$ is equal to 2.

Answer and Step-by-Step Explanation:

Part a)

To find the average value of $f(x)$ on the interval $[0, 2]$, we need to evaluate the definite integral of $f(x)$ on this interval and divide the result by the width of the interval.

The average value of $f(x)$ on $[0, 2]$ is given by:

Now, we can find the integral of $f(x)$ by taking the antiderivative of each term individually:

Now, we add up these integrals to get the overall integral:

Finally, we divide this by the width of the interval (2 - 0 = 2):

Therefore, the average value of the function $f(x) = 2x^3 - 3x^2 + 4$ on the interval $[0, 2]$ is 4.

Part b)

To determine the x-coordinate(s) at which the average value of $f(x)$ on the interval $[0, x]$ is equal to 2, we need to find the value(s) of $x$ that satisfy the following equation:

We already know that the integral of $f(x)$ on the interval $[0, x]$ is given by:

To evaluate this integral, we perform similar steps as in Part a):

Now, we add up these integrals:

Substituting this back into the equation $\frac{1}{x - 0} \int_0^x f(x) \,dx = 2$:

Simplifying:

Rearranging to get a quadratic equation:

Solving this equation for $x$ using numerical methods or factoring techniques, we find that the solutions are approximately $x \approx -0.618$, $x \approx 2.618$, and $x \approx 3.193$.

Therefore, the x-coordinate(s) at which the average value of $f(x)$ on the interval $[0, x]$ is equal to 2 are approximately $x \approx -0.618$, $x \approx 2.618$, and $x \approx 3.193$.