AP Calculus BC Quiz: Average Value Of Functions On Intervals

What this quiz covers

This quiz focuses on Average Value Of Functions On Intervals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{1}{2}\int_{1}^{3} t^2\,dt$
2. $\int_{1}^{3} t^2\,dt$
3. $s(2)$
4. $\dfrac{13}{3}$ (correct answer)
5. $\dfrac{26}{3}$

Explanation: The skill here is calculating the average value of a function over an interval, applied to finding the average speed from a speed function. The formula (1/(b-a)) ∫_a^b f(t) dt gives the average speed, equivalent to total distance divided by time if speed is non-negative. For s(t) = t^2 on [1,3], we use (1/2) ∫_1^3 t^2 dt, evaluating to 13/3 m/s. This reflects the increasing nature of t^2, so the average is between s(1) = 1 and s(3) = 9, specifically weighted by the integral. A tempting distractor is choice E, 26/3, which is the integral without dividing by the interval length of 2. To find the average value, always: identify the interval [a,b], compute ∫_a^b f(t) dt accurately, divide by (b-a), and simplify the result.

Question 2

1. $2$
2. $4$ (correct answer)
3. $6$
4. $8$
5. $10$

Explanation: The skill here involves calculating the average value of a function over an interval, representing the constant power that would deliver the same total energy as the varying output. For the heater's power P(t) = 6 - 2t kW from t = 0 to t = 2, the average is (1/(2 - 0)) ∫_0^2 P(t) dt, with the integral giving total energy in kWh if time is hours, but here it's the accumulation for averaging. Conceptually, this captures the linear decrease in power, finding the equivalent steady level. To apply it, integrate 6t - t² from 0 to 2, getting 8, then divide by 2 to yield 4 kW. A tempting distractor like 8 could result from computing the integral but forgetting to divide by the interval length, representing total instead of average. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

Question 3

1. The value $f(1)$, the midpoint input
2. The total signed area $\int_{-2}^{4} f(x)\,dx$
3. The slope of the secant line from $x=-2$ to $x=4$
4. A constant $c$ such that $\int_{-2}^{4} f(x)\,dx=c(4-(-2))$ (correct answer)
5. The maximum value of $f$ on $[-2,4]$

Explanation: The skill here involves understanding the average value of a function over an interval, which geometrically is the height of a rectangle with width (b - a) and area equal to the integral. For f(x) = 2x - 1 from x = -2 to x = 4, the average value is a constant c such that ∫_{-2}^4 f(x) dx = c(4 - (-2)), directly from the definition. Conceptually, this c represents the mean output level over the linear function's span. While computing gives c = 1, the representation is about equating the integral to c times the length. A tempting distractor like the total signed area might confuse the integral itself with the average, but the average scales it by the interval. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

Question 4

1. $P'(1.5)$
2. $\int_0^3 (6t-2)\,dt$
3. $\dfrac{1}{3}\int_0^3 (6t-2)\,dt$ (correct answer)
4. $\dfrac{1}{2}\int_0^3 (6t-2)\,dt$
5. $\dfrac{P(3)-P(0)}{9}$

Explanation: This problem asks for the average rate of change given $P'(t) = 6t - 2$ on $[0,3]$. The average value of the derivative $P'(t)$ equals the average rate of change of $P(t)$, calculated as $\frac{1}{3-0} \int_0^3 (6t-2) \, dt = \frac{1}{3} \int_0^3 (6t-2) \, dt$. Integrating: $\int (6t-2) \, dt = 3t^2 - 2t$, evaluated from 0 to 3 gives $(27-6) - 0 = 21$. The average is $21/3 = 7$. Choice B (the integral alone) represents the total change $P(3) - P(0)$, not the average rate. When finding average rates: use the average value formula on the derivative, integrate carefully, and divide by the interval length to get the average rate per unit time.

Question 5

1. $\int_{0}^{2}(6-2t)\,dt$
2. $r(1)$
3. $\dfrac{r(0)+r(2)}{2}$
4. $4$ (correct answer)
5. $\dfrac{1}{2}\int_{0}^{2}(6-2t)\,dt$

Explanation: This problem asks for the average inflow rate given r(t) = 6 - 2t on [0,2]. The average value formula gives us (1/(2-0))∫0 to 2dt = (1/2)∫0 to 2dt. Evaluating the integral: [6t - t²] from 0 to 2 = 12 - 4 - 0 = 8. Therefore, the average value is 8/2 = 4. Choice E shows the correct formula setup, while choice B incorrectly uses the function value at the midpoint. Average value represents the constant rate that would produce the same total flow.

Question 6

1. $v(1)$
2. $\dfrac{1}{2}\int_0^2 (6-2t)\,dt$ (correct answer)
3. $\int_0^2 (6-2t)\,dt$
4. $\dfrac{1}{2}\int_0^2 (6+2t)\,dt$
5. $v(2)$

Explanation: This problem asks for the average value of a velocity function over a time interval. The average value formula for a function f(t) on [a,b] is (1/(b-a))∫[a to b] f(t)dt, which gives the mean height of the function over that interval. For velocity v(t) = 6-2t on [0,2], we calculate (1/(2-0))∫0 to 2dt = (1/2)∫0 to 2dt. Choice C represents the total displacement (integral alone), not the average velocity. Choice A gives the velocity at just one moment, missing the averaging process entirely. Remember the average value checklist: determine interval endpoints, apply the formula with correct interval length in denominator, and integrate the given function.

Question 7

1. $p(1)$
2. $\int_0^1 (8-4t)\,dt$
3. $\dfrac{1}{1-0}\int_0^1 (8-4t)\,dt$ (correct answer)
4. $p\!\left(\dfrac{1}{2}\right)$
5. $\dfrac{1}{2}\int_0^1 (8-4t)\,dt$

Explanation: This problem asks for the average rate of population change over a time interval. The average value formula states that for f(t) on [a,b], the average is (1/(b-a))∫[a to b] f(t)dt, representing the constant rate giving the same total change. For rate p(t) = 8-4t on [0,1], we calculate (1/(1-0))∫0 to 1dt = (1/1)∫0 to 1dt = ∫0 to 1dt. Choice B omits the division by interval length, giving total population change instead of average rate. Choice D evaluates the rate at the midpoint, which may not equal the average for non-constant rates. To find average values: determine interval length (here 1), integrate the rate function, then divide by 1.

Question 8

1. The total signed area $\int_{1}^{4}(2x-1)\,dx$
2. The constant height $c$ so $\int_{1}^{4}c\,dx=\int_{1}^{4}m(x)\,dx$ (correct answer)
3. The value at the midpoint, $m\!\left(\dfrac{5}{2}\right)$
4. The endpoint average, $\dfrac{m(1)+m(4)}{2}$
5. The interval length, $4-1$

Explanation: This conceptual problem asks what the average value represents for m(x) = 2x - 1 on [1,4]. The average value is the constant height c such that the rectangle with base [1,4] and height c has the same area as the region under m(x). This means c(4-1) = ∫1 to 4dx, or equivalently, ∫[1 to 4]c dx = ∫[1 to 4]m(x)dx. Choice A gives total area but not average height, while choices C and D give specific function values that don't generally equal the average. Average value = constant function with same integral over the interval.

Question 9

1. $\dfrac{2}{\pi}\int_0^{\pi/2} \cos x\,dx$ (correct answer)
2. $\int_0^{\pi/2} \cos x\,dx$
3. $g\!\left(\dfrac{\pi}{4}\right)$
4. $\dfrac{1}{2}\int_0^{\pi/2} \cos x\,dx$
5. $\dfrac{\pi}{2}\int_0^{\pi/2} \cos x\,dx$

Explanation: This problem requires finding the average value of a trigonometric function. The average value of f(x) on [a,b] equals (1/(b-a))∫[a to b] f(x)dx, which finds the horizontal line with the same area underneath. For g(x) = cos(x) on [0,π/2], the interval length is π/2-0 = π/2, so we need (1/(π/2))∫[0 to π/2] cos x dx = (2/π)∫[0 to π/2] cos x dx. Choice B gives just the integral (total area under cosine), not the average height. Choice C evaluates cosine at one point rather than averaging over the entire interval. Remember: calculate interval length, integrate the function, then divide by interval length—here multiply by 2/π.

Question 10

1. $\dfrac{1}{4}\int_0^4 (2+t)\,dt$ (correct answer)
2. $\int_0^4 (2+t)\,dt$
3. $r(2)$
4. $4$
5. $\dfrac{1}{2}\int_0^4 (2+t)\,dt$

Explanation: This problem requires finding the average value of a rate function over a time interval. The average value of a function f(t) on interval [a,b] is given by (1/(b-a))∫[a to b] f(t)dt, which represents the total accumulated quantity divided by the time elapsed. For the inflow rate r(t) = 2+t on [0,4], we need (1/(4-0))∫0 to 4dt = (1/4)∫0 to 4dt. Choice B gives the total water that flowed in (integral without the 1/4 factor), not the average rate. Choice C evaluates the rate at one instant rather than averaging over the entire interval. To find average values: identify the interval length, set up the integral of the function, then divide the integral by the interval length.

Question 11

1. $\dfrac{1}{e}\int_1^e \dfrac{1}{x}\,dx$
2. $\dfrac{1}{e-1}\int_1^e \dfrac{1}{x}\,dx$ (correct answer)
3. $h\!\left(\dfrac{1+e}{2}\right)$
4. $\int_1^e \dfrac{1}{x}\,dx$
5. $\dfrac{1}{2}\int_1^e \dfrac{1}{x}\,dx$

Explanation: This problem requires identifying the correct average value expression for a function. The average value of f(x) on [a,b] is (1/(b-a))∫[a to b] f(x)dx, which gives the rectangle height having the same area as under the curve. For h(x) = 1/x on [1,e], the interval length is e-1, so the average value equals (1/(e-1))∫1 to edx. Choice D gives just the integral (ln(e)-ln(1) = 1), representing total area without averaging. Choice A incorrectly uses 1/e as the denominator instead of 1/(e-1). For average values: calculate the exact interval length b-a, set up the integral, then divide by that interval length.

Question 12

1. $\dfrac{10}{3}$ (correct answer)
2. $4$
3. $\dfrac{1}{2}\int_0^2 (t^2+2)\,dt$
4. $\int_0^2 (t^2+2)\,dt$
5. $v(1)$

Explanation: This problem requires finding the average value of a velocity function over a time interval. The average value formula states that for a function f on interval [a,b], the average equals (1/(b-a))∫[a to b] f(x)dx. For v(t) = t² + 2 on [0,2], we calculate (1/(2-0))∫[0 to 2](t² + 2)dt = (1/2)[t³/3 + 2t] from 0 to 2 = (1/2)[(8/3 + 4) - 0] = (1/2)(20/3) = 10/3. Choice C shows (1/2)∫[0 to 2](t² + 2)dt, which is tempting but represents only half the integral, not the average value. When finding average values: (1) identify the interval length, (2) set up the integral, (3) multiply by 1/(interval length), and (4) evaluate completely.

Question 13

1. $\dfrac{4}{3}$ (correct answer)
2. $\dfrac{8}{3}$
3. $2$
4. $\dfrac{2}{3}$
5. $4$

Explanation: The skill here involves calculating the average value of a function over an interval, which for a rate of change gives the constant rate that would produce the same total change. For the population rate p(t) = √t thousand per year from t = 0 to t = 4, the average is (1/(4 - 0)) ∫_0^4 √t dt, where the integral computes total population change and dividing averages it. Conceptually, this smooths the concave-up square root function over the interval. To apply it, integrate (2/3)t^{3/2} from 0 to 4, yielding 16/3, then divide by 4 to get 4/3 thousand per year. A tempting distractor like 8/3 might come from mistakenly doubling the integral or confusing with net change, but the correct process divides the integral by the length. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

Question 14

1. $v\!\left(\dfrac{\pi}{2}\right)$
2. $\dfrac{1}{\pi}\int_{0}^{\pi}(5+2\cos t)\,dt$
3. $\int_{0}^{\pi}(5+2\cos t)\,dt$
4. $5$ (correct answer)
5. $5+\dfrac{4}{\pi}$

Explanation: The skill here is calculating the average value of a function over an interval, used to find the average velocity from a velocity function. The average is (1/(b-a)) ∫_a^b f(t) dt, giving the constant velocity for the same displacement. For v(t) = 5 + 2 cos t on [0,π], this is (1/π) ∫_0^π (5 + 2 cos t) dt, equaling 5 m/s since the cosine term integrates to zero over the interval. This shows how periodic variations average out. A tempting distractor is choice E, 5 + 4/π, which might result from incorrectly integrating 2 cos t as 4 sin t or mishandling limits. To find the average value, always: identify the interval [a,b], compute ∫_a^b f(t) dt accurately, divide by (b-a), and simplify the result.

Question 15

1. $\dfrac{1}{e-1}\int_1^e \dfrac{1}{x}\,dx$ (correct answer)
2. $\int_1^e \dfrac{1}{x}\,dx$
3. $\dfrac{1}{\frac{1+e}{2}}$
4. $\dfrac{1}{e-1}$
5. $\dfrac{1}{e+1}\int_1^e \dfrac{1}{x}\,dx$

Explanation: This problem tests finding the average value of $f(x) = \dfrac{1}{x}$ on $[1,e]$. The average value formula gives $\dfrac{1}{e-1} \int_1^e \dfrac{1}{x} \, dx$. The integral of $\dfrac{1}{x}$ is $\ln|x|$, so we evaluate $\ln|e| - \ln|1| = 1 - 0 = 1$. Therefore, the average value is $\dfrac{1}{e-1}$, which matches the expression in choice A. Choice B (just the integral) gives the area under the curve, not the average height. For average value calculations: identify the interval length $(b-a)$, set up the integral with the $\dfrac{1}{b-a}$ factor, integrate using standard formulas, and simplify the final expression.

Question 16

1. $10+4\cos\!\left(\dfrac{\pi}{2}\right)$
2. $\dfrac{1}{\pi}\int_0^{\pi} (10+4\cos t)\,dt$ (correct answer)
3. $\int_0^{\pi} (10+4\cos t)\,dt$
4. $10+4\cos(\pi)$
5. $\dfrac{1}{2}\int_0^{\pi} (10+4\cos t)\,dt$

Explanation: This temperature problem requires finding the average value of T(t) = 10 + 4cos t on [0,π]. The average value formula gives us (1/(π-0))∫[0 to π](10+4cos t)dt = (1/π)∫[0 to π](10+4cos t)dt. Integrating: ∫(10+4cos t)dt = 10t + 4sin t, evaluated from 0 to π gives (10π + 4sin π) - (0 + 4sin 0) = 10π + 0 - 0 = 10π. Therefore, the average temperature is 10π/π = 10°C. Choice C (the integral without 1/π) represents the total accumulated temperature-time, not the average. The average value process requires: setting up (1/(b-a))∫f(x)dx, integrating term by term, evaluating at endpoints, and dividing by the interval length.

Question 17

1. $\dfrac{1}{3}\int_{-1}^{2} (x^2+1)\,dx$ (correct answer)
2. $\int_{-1}^{2} (x^2+1)\,dx$
3. $g\!\left(\dfrac{-1+2}{2}\right)$
4. $\dfrac{1}{2}\int_{-1}^{2} (x^2+1)\,dx$
5. $\dfrac{1}{1}\int_{-1}^{2} (x^2+1)\,dx$

Explanation: This problem tests recognizing the correct average value expression for g(x) = x² + 1 on [-1,2]. The average value formula requires dividing by the interval length: (2-(-1)) = 3, giving (1/3)∫-1 to 2dx. This matches choice A exactly. Choice C (evaluating at the midpoint) only works for linear functions, not for x² + 1 which is quadratic. Choice B (the integral alone) gives the area, not the average height. The average value checklist: compute interval length b-a carefully (especially with negative endpoints), multiply the integral by 1/(b-a), and never confuse total area with average value.

Question 18

1. $\dfrac{6}{\pi}$
2. $3$ (correct answer)
3. $4$
4. $3+\cos\!\left(\dfrac{\pi}{2}\right)$
5. $3\pi$

Explanation: The skill here involves calculating the average value of a function over an interval, which for speed gives the constant velocity matching the total distance traveled. For the cyclist's speed s(t) = 3 + cos t m/s from t = 0 to t = π, the average is (1/π) ∫_0^π s(t) dt, integrating the oscillatory addition to the base speed. Conceptually, the cosine averages to zero over a half-period, leaving the constant term dominant. To apply it, integrate 3t + sin t from 0 to π, getting 3π, then divide by π to obtain 3 m/s. A tempting distractor like 3π might result from confusing total distance with average speed, forgetting to divide by time. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

Question 19

1. $\dfrac{15}{4}$
2. $\dfrac{5}{4}$ (correct answer)
3. $\dfrac{5}{3}$
4. $\dfrac{1}{3}$
5. $\dfrac{15}{3}$

Explanation: The skill here involves calculating the average value of a function over an interval, accounting for both positive and negative areas if present. For f(x) = x^3 from x = -1 to x = 2, the average is (1/(2 - (-1))) ∫_{-1}^2 x^3 dx, where the integral captures the net area under the odd-powered curve. Conceptually, the negative area from -1 to 0 partially cancels the positive from 0 to 2, affecting the mean. To apply it, evaluate (1/4)x^4 from -1 to 2, getting 15/4, then divide by 3 to yield 5/4. A tempting distractor like 15/4 might come from omitting the division by interval length, representing net area instead of average. For any average value problem, checklist: identify the interval [a, b], compute the definite integral of the function, divide by (b - a), and interpret as the mean level over that span.

Question 20

1. $0$
2. $\dfrac{2}{\pi}$ (correct answer)
3. $\dfrac{1}{\pi}\int_{0}^{\pi}\sin x\,dx$
4. $\int_{0}^{\pi}\sin x\,dx$
5. $g\!\left(\dfrac{\pi}{2}\right)$

Explanation: This problem asks for the average value of g(x) = sin x on [0,π]. Using the average value formula: (1/(π-0))∫[0 to π] sin x dx = (1/π)∫[0 to π] sin x dx. The integral evaluates to [-cos x] from 0 to π = -(-1) - (-1) = 2. Therefore, the average value is 2/π. Choice C shows the formula but doesn't evaluate it, while choice E incorrectly uses the function value at the midpoint. For average values: set up the formula, evaluate the integral, then divide by interval length.