AP Calculus AB: Average Value and Motion Applications (Drill 1)

Question 1. The average value of \( f(x) = 6\sqrt{x} \) on the interval \( [0, 4] \) is:

• A) \( 8 \) ✓
• B) \( 32 \)
• C) \( 6 \)
• D) \( 12 \)

Explanation: Choice A is correct. The average value formula is \( \dfrac{1}{b-a}\int_a^b f(x)\,dx \). Here, \( \dfrac{1}{4}\int_0^4 6x^{1/2}\,dx = \dfrac{1}{4}\left[4x^{3/2}\right]_0^4 = \dfrac{1}{4}(32) = 8 \). Choice B is incorrect because the student computed the definite integral \( \int_0^4 6\sqrt{x}\,dx = 32 \) correctly but forgot to divide by \( b - a = 4 \). Choice C is incorrect because the student computed the arithmetic mean of the endpoint values, \( \dfrac{f(0)+f(4)}{2} = \dfrac{0+12}{2} = 6 \), confusing the average value of a function over an interval with the average of two output values. Choice D is incorrect because the student found the antiderivative of \( 6\sqrt{x} \) as \( 3x^2 \) (treating \( \sqrt{x} \) as \( x \) rather than \( x^{1/2} \)), yielding \( \dfrac{1}{4}\left[3x^2\right]_0^4 = \dfrac{48}{4} = 12 \).

Question 2. Let \( v(t) = t^2 \) on the interval \( [0, 3] \). The Mean Value Theorem for Integrals guarantees a value \( c \) in \( (0, 3) \) such that \( v(c) \) equals the average value of \( v \) on \( [0, 3] \). What is the value of \( c \)?

• A) \( \sqrt{3} \) ✓
• B) \( 3 \)
• C) \( \dfrac{3}{2} \)
• D) \( \dfrac{3\sqrt{2}}{2} \)

Explanation: Choice A is correct. First, find the average value: \( \dfrac{1}{3}\int_0^3 t^2\,dt = \dfrac{1}{3}\left[\dfrac{t^3}{3}\right]_0^3 = \dfrac{1}{3}(9) = 3 \). Then set \( v(c) = 3 \): \( c^2 = 3 \), so \( c = \sqrt{3} \approx 1.73 \), which lies in \( (0, 3) \). Choice B is incorrect because the student omitted the \( \dfrac{1}{b-a} \) factor, computed the full integral as the "average," set \( c^2 = 9 \), and obtained \( c = 3 \), an endpoint, not an interior value. Choice C is incorrect because the student applied the Mean Value Theorem for derivatives rather than for integrals, setting \( v'(c) = \dfrac{v(3)-v(0)}{3-0} = 3 \), so \( 2c = 3 \) and \( c = \dfrac{3}{2} \). Choice D is incorrect because the student computed the average of the endpoint values \( \dfrac{v(0)+v(3)}{2} = \dfrac{0+9}{2} = 4.5 \), set \( c^2 = 4.5 \), and obtained \( c = \dfrac{3\sqrt{2}}{2} \), confusing the average of outputs with the true average value of the function.

Question 3. A particle moves along the x-axis with velocity \( v(t) \) feet per second. On the interval \( [0, 5] \), the velocity is negative for \( 0 < t < 3 \) and positive for \( 3 < t < 5 \). Which of the following expressions gives the total distance traveled by the particle over \( [0, 5] \)?

• A) \( \displaystyle\int_0^5 v(t)\,dt \)
• B) \( \displaystyle\left|\int_0^5 v(t)\,dt\right| \)
• C) \( \displaystyle -\int_0^3 v(t)\,dt + \int_3^5 v(t)\,dt \) ✓
• D) \( \displaystyle \int_0^3 v(t)\,dt - \int_3^5 v(t)\,dt \) over the stated interval

Explanation: Choice C is correct. Total distance is \( \int_0^5 |v(t)|\,dt \). Since \( v(t) 0 \) on \( (3,5) \), \( |v(t)| = v(t) \). So total distance \( = -\int_0^3 v(t)\,dt + \int_3^5 v(t)\,dt \). Choice A is incorrect because \( \int_0^5 v(t)\,dt \) gives displacement, the signed net change in position, not total distance; the negative and positive portions partially cancel. Choice B is incorrect because taking the absolute value of the net displacement is not the same as summing the actual distances on each piece. For example, if a particle travels 5 feet left and then 3 feet right, the net displacement is \( -2 \) feet, with absolute value 2, but the total distance is 8 feet. Choice D is incorrect because the student negated the wrong piece, subtracting the positive-velocity portion and leaving the negative-velocity portion as-is, which would yield a negative result.

Question 4. A particle moves along a straight line with velocity \( v(t) = t^2 - 6t + 8 \) feet per second for \( 0 \le t \le 5 \). What is the total distance traveled by the particle?

• A) \( \dfrac{28}{3} \) ✓
• B) \( \dfrac{20}{3} \)
• C) \( 8 \)
• D) \( \dfrac{16}{3} \)

Explanation: Choice A is correct. Factor: \( v(t) = (t-2)(t-4) \). The velocity is zero at \( t = 2 \) and \( t = 4 \), positive on \( [0,2) \), negative on \( (2,4) \), and positive on \( (4,5] \). Using antiderivative \( F(t) = \dfrac{t^3}{3} - 3t^2 + 8t \): \( F(0) = 0 \), \( F(2) = \dfrac{20}{3} \), \( F(4) = \dfrac{16}{3} \), \( F(5) = \dfrac{20}{3} \). Total distance is the sum of the absolute values of displacement on each subinterval: \( |F(2)-F(0)| + |F(4)-F(2)| + |F(5)-F(4)| = \dfrac{20}{3} + \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{28}{3} \). Choice B is incorrect because the student computed net displacement \( F(5)-F(0) = \dfrac{20}{3} \) without splitting the integral at the zeros of \( v(t) \); displacement and total distance differ whenever the particle changes direction. Choice C is incorrect because the student correctly found the zero at \( t = 2 \) but missed the second zero at \( t = 4 \), summing only two distance pieces: \( \dfrac{20}{3} + \dfrac{4}{3} = 8 \). Choice D is incorrect because the student stopped at \( t = 4 \), the second zero of velocity, and computed \( F(4)-F(0) = \dfrac{16}{3} \), omitting the distance traveled on \( [4, 5] \).

Question 5. A particle moves along the x-axis with velocity \( v(t) = 2t + 1 \) feet per second. If the particle's position at \( t = 0 \) is \( x(0) = 3 \), what is the particle's position at \( t = 4 \)?

• A) \( 23 \) ✓
• B) \( 20 \)
• C) \( 9 \)
• D) \( 19 \)

Explanation: Choice A is correct. Position is found by integrating velocity and applying the initial condition: \( x(4) = x(0) + \int_0^4 v(t)\,dt = 3 + \int_0^4 (2t+1)\,dt = 3 + \left[t^2+t\right]_0^4 = 3 + 20 = 23 \). Choice B is incorrect because the student correctly computed the displacement \( \int_0^4(2t+1)\,dt = 20 \) but forgot to add the initial position \( x(0) = 3 \); displacement and position are not the same thing. Choice C is incorrect because the student evaluated the velocity at \( t = 4 \), finding \( v(4) = 9 \), rather than integrating to find the accumulated change in position. Choice D is incorrect because the student dropped the constant term when finding the antiderivative, using \( t^2 \) instead of \( t^2 + t \), and computed \( \int_0^4 2t\,dt = 16 \), giving \( 3 + 16 = 19 \).