Average Value of Functions on Intervals - AP Calculus BC

For $g(x)=x^2+1$ on $[-1,2]$ , which expression equals the average value of $g$ on the interval?
Answer: $\dfrac{1}{3}\int_{-1}^{2} (x^2+1)\,dx$ 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 2](x²+1)dx. 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.
A population model is $p(t)=\frac{1}{t+1}$ for $0\le t\le 1$ ; what is the average value of $p$ on $[0,1]$ ?
Answer: $\ln 2$ This problem involves finding the average value of a function over a given interval, which is a fundamental application of definite integrals in calculus. The formula for the average value of p(t) on [0,1] is (1/(1-0)) ∫_0^1 1/(t+1) dt. Compute the integral: the antiderivative is ln|t+1|, evaluated as ln2 - ln1 = ln2 - 0 = ln2. Dividing by 1 gives ln2, representing the average population over the time. A tempting distractor is 1/2, perhaps from evaluating the function at the midpoint t=0.5 as 1/1.5=2/3, but close yet incorrect without integration. To compute average values reliably, always: identify the function and interval, find the antiderivative, evaluate the definite integral, divide by (b-a), and check if the result lies between the function's minimum and maximum.
A particle's speed is $s(t)=\sqrt{t}$ for $0\le t\le1$ ; what is the average speed over this interval?
Answer: $\dfrac{1}{1}\int_0^1 \sqrt{t}\,dt$ This problem requires finding the average speed over a time interval. The average value of a function f(t) on [a,b] is (1/(b-a))∫[a to b] f(t)dt, which gives the constant speed that would cover the same distance. For speed s(t) = √t on [0,1], we calculate (1/(1-0))∫[0 to 1]√t dt = (1/1)∫[0 to 1]√t dt = ∫[0 to 1]√t dt. Choice A omits the division by interval length, giving total distance instead of average speed. Choice B evaluates speed at one instant rather than averaging over the entire trip. The average value process: find interval length (here it's 1), integrate the function, then divide by interval length to get the mean value.
Let $F(x)=\cos x$ on $\left[0,\dfrac{\pi}{2}\right]$ . What is the average value of $F$ on this interval?
Answer: $\dfrac{2}{\pi}$ This problem requires finding the average value of F(x) = cos x on [0,π/2]. Using the average value formula: (1/(π/2-0))∫[0 to π/2] cos x dx = (2/π)∫[0 to π/2] cos x dx. The integral evaluates to [sin x] from 0 to π/2 = 1 - 0 = 1. Therefore, the average value is (2/π)(1) = 2/π. Choice B shows the formula but doesn't evaluate it, while choice E incorrectly uses the arithmetic mean of endpoints. For trigonometric functions, average value rarely equals endpoint averages.
If a car's velocity is $v(t)=3t^2-6t+4$ (m/s) for $0\le t\le2$ , what is its average velocity?
Answer: $\dfrac{1}{2}\int_{0}^{2}v(t)\,dt$ This problem tests your understanding of average value of a function, specifically finding average velocity from a velocity function. The average value of a continuous function f on interval [a,b] is given by (1/(b-a))∫[a to b]f(x)dx. For velocity v(t) on [0,2], the average velocity is (1/(2-0))∫[0 to 2]v(t)dt = (1/2)∫[0 to 2]v(t)dt. Choice C might tempt you because it's the total displacement, but average velocity requires dividing by the time interval. To find average values: identify the interval length (b-a), compute the integral, then divide the integral by the interval length.
For $h(x)=\dfrac{1}{x}$ on $[1,e]$ , which expression equals the average value of $h$ on this interval?
Answer: $\dfrac{1}{e-1}\int_1^e \dfrac{1}{x}\,dx$ 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 e](1/x)dx. 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.
For $f(x)=\sin x$ on $[0,\pi]$ , what is the average value of $f$ on the interval?
Answer: $\dfrac{2}{\pi}$ This problem requires finding the average value of the sine function over one complete arch. The average value of f(x) = sin x on [0,π] equals (1/(π-0))∫[0 to π]sin x dx. The integral ∫sin x dx = -cos x, so we evaluate [-cos x] from 0 to π = -cos(π) - (-cos(0)) = -(-1) - (-1) = 2. Therefore, the average is (1/π)(2) = 2/π. Choice E suggests f(π/2) = sin(π/2) = 1, which is the maximum value, not the average—a common misconception that the average occurs at the midpoint. For trigonometric averages: (1) set up the average value integral, (2) use antiderivative formulas, (3) evaluate at endpoints carefully, and (4) remember average ≠ value at midpoint.
A particle's position is $x(t)=t^3$ (m) for $0\le t\le2$ . What is the average velocity on $[0,2]$ ?
Answer: $\dfrac{1}{2}\int_0^2 x'(t)\,dt$ For a particle with position x(t) = t³, the average velocity on [0,2] equals the average value of the velocity function x'(t). Since x'(t) = 3t², the average velocity is (1/(2-0))∫[0 to 2]x'(t)dt = (1/2)∫[0 to 2]3t²dt. By the Fundamental Theorem, ∫[0 to 2]x'(t)dt = x(2) - x(0) = 8 - 0 = 8, so the average velocity is 8/2 = 4 m/s. Choice A incorrectly divides the displacement by 2 without recognizing this equals (1/2)∫x'(t)dt. For average velocity: use the average value of x'(t), which equals (x(b)-x(a))/(b-a) by the Fundamental Theorem, giving a computational shortcut.
For $f(x)=\dfrac{1}{x}$ on $[1,e]$ , what is the average value of $f$ on this interval?
Answer: $\dfrac{1}{e-1}\int_1^e \dfrac{1}{x}\,dx$ 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.
A population changes at rate $P'(t)=6t-2$ for $0 \le t \le 3$ . What is the average rate of change on $[0,3]$ ?
Answer: $\dfrac{1}{3}\int_0^3 (6t-2)\,dt$ 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.
For $g(x)=4-x$ on $[0,4]$ , what is the average value of $g$ on the interval?
Answer: $\dfrac{1}{4}\int_{0}^{4}(4-x)\,dx$ This problem tests finding the average value of a linear function on its domain. For g(x) = 4-x on [0,4], the average value is (1/(4-0))∫[0 to 4](4-x)dx = (1/4)∫[0 to 4](4-x)dx. Choice D gives (g(0)+g(4))/2 = (4+0)/2 = 2, which happens to equal the correct average for this linear function, but this endpoint averaging method doesn't work for non-linear functions. Always use the integral formula: average = (1/(b-a))∫[a to b]f(x)dx for guaranteed correctness.
For $f(x)=x^2+1$ on $[-1,1]$ , which expression equals the average value of $f$ on the interval?
Answer: $\dfrac{1}{1-(-1)}\int_{-1}^{1} (x^2+1)\,dx$ This problem asks for the expression representing the average value of a function on an interval. The average value formula states that for f(x) on [a,b], the average equals (1/(b-a))∫[a to b] f(x)dx, which finds the rectangle height with the same area as under the curve. For f(x) = x²+1 on [-1,1], the interval length is 1-(-1) = 2, so we need (1/(1-(-1)))∫[-1 to 1](x²+1)dx. Choice C gives just the integral (total area), not the average height. Choice B incorrectly uses 1/2 instead of 1/2 as the denominator. To find average values correctly: calculate b-a for interval length, set up the integral, then divide by the interval length.
A device's power use is $p(t)=4-2t$ (W) for $0 \le t \le 1$ . What is the average power on $[0,1]$ ?
Answer: $\dfrac{1}{1-0}\int_0^1 (4-2t)\,dt$ Finding the average power requires applying the average value formula to $p(t) = 4 - 2t$ on $[0,1]$ . The average value is $\frac{1}{1-0} \int_0^1 (4-2t)\, dt = \int_0^1 (4-2t)\, dt$ since the interval length is 1. Integrating: $\int (4-2t)\, dt = 4t - t^2$ , evaluated from 0 to 1 gives $(4-1) - 0 = 3$ W. Choice D (multiplying by $\frac{1}{2}$ instead of $\frac{1}{1}$ ) is incorrect because the interval length is 1, not 2. For average value problems: always identify the correct interval length $(b-a)$ , set up $\frac{1}{b-a} \int f(t)\, dt$ , and remember that when $b-a = 1$ , the average equals the integral value.
Let $F(x)=2-\sqrt{x}$ for $0\le x\le 4$ . What is the average value of $F$ on $[0,4]$ ?
Answer: $\dfrac{2}{3}$ This problem requires finding the average value of a function involving a square root. For F(x) = 2 - √x on [0,4], the average equals (1/(4-0))∫[0 to 4](2 - √x)dx = (1/4)∫[0 to 4](2 - x^(1/2))dx. Computing: ∫(2 - $x^{1/2}$ )dx = 2x - (2/3) $x^{3/2}$ , so we evaluate [2x - (2/3) $x^{3/2}$ ] from 0 to 4 = [8 - (2/3)(8)] - [0] = 8 - 16/3 = 24/3 - 16/3 = 8/3. Therefore, the average is (1/4)(8/3) = 8/12 = 2/3. Choice C shows (1/4)∫[0 to 4](2 - √x)dx, which is the correct setup but not the evaluated answer. For functions with roots: (1) rewrite roots as fractional exponents, (2) use power rule for integration, (3) evaluate carefully at endpoints, and (4) simplify fractions completely.
Let $g(x)=3x-1$ for $1\le x\le 3$ . What is the average value of $g$ on $[1,3]$ ?
Answer: $5$ This problem asks for the average value of a linear function on a given interval. For g(x) = 3x - 1 on [1,3], the average equals (1/(3-1))∫[1 to 3](3x - 1)dx = (1/2)∫[1 to 3](3x - 1)dx. Computing: ∫(3x - 1)dx = (3x²/2) - x, so we evaluate [(3x²/2) - x] from 1 to 3 = [(27/2) - 3] - [(3/2) - 1] = (21/2) - (1/2) = 10. Therefore, the average is (1/2)(10) = 5. Choice D shows (1/2)∫[1 to 3](3x - 1)dx, which represents the calculation but not the final evaluated answer. For linear function averages: (1) apply the standard formula, (2) integrate term by term, (3) evaluate at both endpoints, and (4) simplify the arithmetic carefully.
A machine's power use is $p(t)=4+\dfrac{t}{2}$ (kW) for $2\le t\le 6$ ; what is the average power?
Answer: $6$ The skill here is calculating the average value of a function over an interval, applied to finding the average power from a power function. The average is (1/(b-a)) ∫_a^b f(t) dt, giving the constant power for the same total energy. For p(t) = 4 + t/2 on [2,6], with b-a=4, it's (1/4) ∫_2^6 (4 + t/2) dt, simplifying to 6 kW. This accounts for the linear increase from 5 to 7. A tempting distractor is choice B, the integral without dividing by 4, yielding total instead of average. 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.
A machine's output rate is $q(t)=4+2\sin t$ units/min for $0\le t\le 2\pi$ ; find the average output rate.
Answer: $4$ The skill here involves calculating the average value of a function over an interval, giving the constant rate for equivalent total output. For the machine's rate q(t) = 4 + 2 sin t from t = 0 to t = 2π, the average is (1/(2π)) ∫_0^{2π} q(t) dt, integrating over a full sine period. Conceptually, the sine term averages to zero, leaving the base rate. To apply it, evaluate 4t - 2 cos t from 0 to 2π, getting 8π, then divide by 2π to yield 4 units per minute. A tempting distractor like 8π might come from taking the integral as the average, forgetting to divide by the interval 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.
A cyclist's speed is $s(t)=3+\cos t$ m/s for $0\le t\le \pi$ ; what is the average speed?
Answer: $3$ 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.
If $f(x)=\dfrac{1}{x}$ on $[1,e]$ , which expression gives the average value of $f$ on $[1,e]$ ?
Answer: $\dfrac{1}{e-1}\int_{1}^{e}\dfrac{1}{x}\,dx$ This question asks for the average value of the reciprocal function on [1,e]. Applying the average value formula to f(x) = 1/x on [1,e], we get (1/(e-1))∫[1 to e](1/x)dx. Choice B incorrectly uses 1/e as the coefficient, dividing by e instead of the interval length (e-1). For any average value problem: first find the interval length by subtracting endpoints, integrate the function, then divide the integral by that interval length.
Electric power is $P(t)=2+\cos t$ (kW) for $0\le t\le2\pi$ . What is the average power used?
Answer: $\dfrac{1}{2\pi}\int_{0}^{2\pi}P(t)\,dt$ This problem involves finding the average power consumption over a complete period of a periodic function. For P(t) on [0,2π], the average value is (1/(2π-0))∫[0 to 2π]P(t)dt = (1/2π)∫[0 to 2π]P(t)dt. Choice B gives just the integral, which represents total energy used rather than average power. The average value process: calculate interval length (2π here), integrate the function, then multiply by 1/(interval length) to get the average.

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