Average Value of Functions on Intervals - AP Calculus AB

For $f(x)=3x+1$ on $[2,6]$ , what is the average value of $f$ on the interval?
Answer: $13$ This problem requires finding the average value of the linear function f(x). The average value of f(x) = 3x + 1 on [2,6] is (1/4)∫[2 to 6] (3x + 1)dx = (1/4)[3x²/2 + x] from 2 to 6 = (1/4)[(54 + 6) - (6 + 2)] = (1/4)(52) = 13. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 13.
A ball's height is $h(t)=16t-4t^2$ for $0\le t\le 4$ . What is the average height on $[0,4]$ ?
Answer: $\dfrac{32}{3}$ This question asks for the average value of the height function h(t). The average value of h(t) = 16t - 4t² on [0,4] is (1/4)∫[0 to 4] (16t - 4t²)dt = (1/4)[8t² - 4t³/3] from 0 to 4 = (1/4)(128 - 256/3) = (1/4)(128/3) = 32/3. Choice D shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 32/3.
For $f(x)=\sin x+2$ on $[0,2\pi]$ , what is the average value of $f$ on the interval?
Answer: $2$ This problem requires finding the average value of $f(x) = \sin(x) + 2$ on $[0, 2\pi]$ . The average value is $\frac{1}{2\pi} \int_0^{2\pi} (\sin(x) + 2) \, dx = \frac{1}{2\pi} [-\cos(x) + 2x]_0^{2\pi} = \frac{1}{2\pi} (0 + 4\pi) = 2$ . Since $\sin(x)$ integrates to zero over a complete period, only the constant term 2 contributes to the average. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result A = 2.
A medication level is $M(t)=4+2t$ (mg/L) for $0\le t\le 6$ . What is the average level on $[0,6]$ ?
Answer: $10$ This question asks for the average value of the medication level function M(t). The average value of M(t) = 4 + 2t on [0,6] is (1/6)∫[0 to 6] (4 + 2t)dt = (1/6)[4t + t²] from 0 to 6 = (1/6)(24 + 36) = 10 mg/L. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 10.
A company's profit rate is $p(t)=12-3t$ (thousand $/\text{day}$ ) for $0\le t\le2$ . What is the average profit rate?
Answer: $9$ This problem asks for the average profit rate over time, requiring you to find the average value of a decreasing linear function. The average of p(t)=12-3t on [0,2] equals (1/2)∫[0 to 2](12-3t)dt. Computing: (1/2)[12t-3t²/2] from 0 to 2 = (1/2)[(24-6)-(0)] = (1/2)(18) = 9 thousand dollars per day. A tempting error is to evaluate p(2)=12-6=6 (choice D), mistaking the endpoint value for the average. To find average values of linear functions: (1) integrate term by term, (2) evaluate at both bounds, (3) remember to multiply by 1/(interval length).
Water flows at rate $R(t)=6-t$ (L/min) for $0\le t\le4$ . What is the average flow rate on $[0,4]$ ?
Answer: $4$ This problem asks for the average value of a rate function over time. The average value of R(t) = 6 - t on [0,4] is found using (1/(4-0))∫[0 to 4](6-t)dt. Evaluating: (1/4)[6t - t²/2] from 0 to 4 = (1/4)[24 - 8] = (1/4)(16) = 4 L/min. Choice C represents the average value formula correctly but unevaluated, while choice B gives the total volume without averaging. Remember the average value checklist: divide the integral by the interval length to get the true average, not just the total accumulation.
For $f(x)=4-x$ on $[-1,3]$ , what is the average value of $f$ on the interval?
Answer: $3$ This problem requires finding the average value of f(x) = 4 - x on [-1,3]. The average value is (1/4)∫[-1 to 3] (4 - x)dx = (1/4)[4x - x²/2] from -1 to 3 = (1/4)[(12 - 4.5) - (-4 - 0.5)] = (1/4)(12) = 3. Choice C shows the correct formula setup but doesn't evaluate it. Choice A represents the average of endpoint values, which equals the average for linear functions but isn't the general approach. The correct answer is the numerical result E = 3.
For $f(x)=\dfrac{x}{1+x}$ on $[0,2]$ , which expression gives the average value of $f$ on $[0,2]$ ?
Answer: $\dfrac{1}{2}\int_0^2 \dfrac{x}{1+x}\,dx$ This problem asks for the correct expression representing the average value of f(x) = x/(1+x) on [0,2]. The average value of any function f(x) on interval [a,b] is given by the formula (1/(b-a))∫[a to b] f(x)dx. For the interval [0,2], this becomes (1/(2-0))∫[0 to 2] f(x)dx = (1/2)∫[0 to 2] f(x)dx. Choice E incorrectly uses 1/4 as the coefficient instead of 1/2. The correct expression is choice A, which properly applies the average value formula.
If $g(x)=x^2$ on $[0,3]$ , what is the average value of $g$ on $[0,3]$ ?
Answer: $3$ This problem requires finding the average value of g(x) = x² on [0,3]. The average value formula gives (1/3)∫[0 to 3] x²dx = (1/3)[x³/3] from 0 to 3 = (1/3)(9) = 3. Choice C shows the correct formula setup but doesn't evaluate it. Choice A represents the midpoint value, which is not generally equal to the average value for nonlinear functions. The correct answer is the numerical result E = 3.
For $f(x)=e^x$ on $[0,1]$ , which expression equals the average value of $f$ on $[0,1]$ ?
Answer: $\dfrac{1}{1-0}\int_0^1 e^x\,dx$ This problem asks for the correct expression representing the average value of f(x) = $e^x$ on [0,1]. The average value of any function f(x) on interval [a,b] is given by the formula (1/(b-a))∫[a to b] f(x)dx. For the interval [0,1], this becomes (1/(1-0))∫[0 to 1] $e^x$ dx = (1/1)∫[0 to 1] $e^x$ dx. Choice E incorrectly includes an extra factor of 1/2 in the denominator. The correct expression is choice A, which properly applies the average value formula.
For $f(x)=x^3$ on $[-1,1]$ , what is the average value of $f$ on the interval?
Answer: $0$ This problem requires finding the average value of f(x) = x³ on [-1,1]. The average value is (1/2)∫[-1 to 1] x³ dx = (1/2)[x⁴/4] from -1 to 1 = (1/2)(1/4 - 1/4) = 0. Since x³ is an odd function and we're integrating over a symmetric interval, the integral equals zero. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 0.
For $f(x)=\sqrt{x}$ on $[0,4]$ , what is the average value of $f$ on the interval?
Answer: $\dfrac{4}{3}$ This problem requires finding the average value of f(x) = √x on [0,4]. The average value is (1/4)∫[0 to 4] √x dx = (1/4)∫[0 to 4] $x^{1/2}$ dx = (1/4)[ $2x^{3/2}$ /3] from 0 to 4 = (1/4)(16/3) = 4/3. Choice A shows the correct formula setup but doesn't evaluate it. Choice C represents the midpoint value, which is not equal to the average value for this nonlinear function. The correct answer is the numerical result E = 4/3.
A population changes at rate $r(t)=3e^{t}$ (thousand/yr) for $0\le t\le1$ . What is the average rate on $[0,1]$ ?
Answer: $3(e-1)$ This problem involves finding the average rate of population change. The average value of r(t) = $3e^t$ on [0,1] is (1/1)∫[0 to 1] $3e^t$ dt = 3[ $e^t$ ] from 0 to 1 = 3(e - 1) thousand/yr. Choice D correctly shows the average value formula with the interval length explicitly written, while choice A gives just the integral without averaging. The exponential function's average is not simply its value at any particular point like t = 1. For average values with exponentials: integrate carefully, apply limits to get $e^b$ - $e^a$ , then divide by interval length.
A signal is $s(t)=\sin(2t)$ for $0\le t\le \pi$ . What is the average value of $s$ on $[0,\pi]$ ?
Answer: $0$ This problem requires finding the average value of s(t) = sin(2t) on [0,π]. The average value is (1/π)∫[0 to π] sin(2t)dt = (1/π)[-cos(2t)/2] from 0 to π = (1/π)[(-1/2) - (-1/2)] = 0. Since sin(2t) completes exactly two full cycles over [0,π], its integral equals zero. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 0.
Temperature is $T(t)=20+4\cos t$ (°C) for $0\le t\le\dfrac{\pi}{2}$ . What is the average temperature?
Answer: $20+\dfrac{8}{\pi}$ This problem tests finding the average value of a temperature function involving cosine, requiring careful integration of trigonometric functions. The average temperature equals (1/(π/2))∫[0 to π/2](20+4cos t)dt = (2/π)∫[0 to π/2](20+4cos t)dt. Computing: (2/π)[20t+4sin t] from 0 to π/2 = (2/π)[(10π+4·1)-(0+0)] = (2/π)(10π+4) = 20+8/π. Students might select C by evaluating T(π/4), mistakenly believing the midpoint value equals the average. To find average values with trig functions: (1) carefully set up with correct interval length, (2) remember derivative/integral pairs for sine and cosine, (3) evaluate trig functions at special angles.
For $f(x)=\cos x$ on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ , what is the average value of $f$ ?
Answer: $\dfrac{2}{\pi}$ This problem requires finding the average value of $f(x) = \cos(x)$ on $[-\pi/2, \pi/2]$ . The average value is $(1/\pi)\int_{-\pi/2}^{\pi/2} \cos(x)\,dx = (1/\pi)[\sin(x)]_{-\pi/2}^{\pi/2} = (1/\pi)(1 - (-1)) = 2/\pi$ . Choice A shows the correct formula setup but doesn't evaluate it. Choice B represents the midpoint value $f(0) = 1$ , which is not equal to the average value for this function. The correct answer is the numerical result E = $2/\pi$ .
Electric current is $I(t)=2+\sqrt{t}$ (A) for $0\le t\le4$ . What is the average current on $[0,4]$ ?
Answer: $\dfrac{10}{3}$ This problem asks for the average current over a time interval. For I(t) = 2 + √t on [0,4], the average value is (1/4)∫[0 to 4](2 + √t)dt = (1/4)[2t + (2/3) $t^{3/2}$ ] from 0 to 4 = (1/4)[8 + (2/3)(8)] = (1/4)[8 + 16/3] = (1/4)(40/3) = 10/3 A. Choice C correctly shows the average value formula, while choice B = 2 + √2 incorrectly evaluates the function at t = 2. Average value checklist for functions with roots: integrate using power rule with fractional exponents, evaluate limits carefully, then divide by interval length.
For $f(x)=x^2$ on $[0,2]$ , what is the average value of $f$ on the interval?
Answer: $\dfrac{4}{3}$ This problem requires finding the average value of a quadratic function. For f(x) = x² on [0,2], the average value equals (1/2)∫[0 to 2]x²dx = (1/2)[x³/3] from 0 to 2 = (1/2)(8/3) = 4/3. Choice B shows the correct average value formula, while choice C gives the total area under the curve without dividing by the interval length. Unlike linear functions, quadratic functions don't average to their midpoint value—here f(1) = 1 ≠ 4/3. Average value checklist: set up (1/(b-a))∫f(x)dx, integrate, evaluate at limits, and simplify the fraction.
A spring force is $F(x)=5x$ for $0\le x\le 4$ . What is the average force on $[0,4]$ ?
Answer: $10$ This question asks for the average value of the force function F(x). The average value of F(x) = 5x on [0,4] is (1/4)∫[0 to 4] 5x dx = (1/4)[5x²/2] from 0 to 4 = (1/4)(40) = 10. Choice A shows the correct formula setup but doesn't evaluate it. Choice B represents the midpoint value, which equals the average for this linear function but isn't the general approach. The correct answer is the numerical result E = 10.
A runner's pace is $p(t)=8-\dfrac{t}{2}$ (min/mi) for $0\le t\le6$ . What is the average pace on $[0,6]$ ?
Answer: $\dfrac{13}{2}$ This problem requires finding the average pace over a time interval. The average value of p(t) = 8 - t/2 on [0,6] is (1/6)∫[0 to 6](8 - t/2)dt. Computing: (1/6)[8t - t²/4] from 0 to 6 = (1/6)[48 - 9] = (1/6)(39) = 13/2 min/mi. Choice C shows the correct formula but unevaluated, while choice D = 8 incorrectly uses just the initial pace value. To find average values of linear functions: the average always equals the function value at the midpoint, which here is p(3) = 8 - 3/2 = 13/2.

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