Average Value of Functions on Intervals
Help Questions
AP Calculus AB › Average Value of Functions on Intervals
For $f(x)=4-x$ on $-1,3$, what is the average value of $f$ on the interval?
$\int_{-1}^3 f(x),dx$
$f(1)$
$3$
$\dfrac{f(-1)+f(3)}{2}$
$\dfrac{1}{4}\int_{-1}^3 f(x),dx$
Explanation
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 $f(x)=3x+1$ on $2,6$, what is the average value of $f$ on the interval?
$\dfrac{1}{4}\int_2^6 f(x),dx$
$13$
$\dfrac{f(2)+f(6)}{2}$
$f(4)$
$\int_2^6 f(x),dx$
Explanation
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 particle’s position is $s(t)=t^2+2t$ for $0\le t\le 2$. What is the average value of $s$ on $0,2$?
$\dfrac{s(0)+s(2)}{2}$
$s(1)$
$\dfrac{1}{2}\int_0^2 s(t),dt$
$\int_0^2 s(t),dt$
$\dfrac{10}{3}$
Explanation
This question requires finding the average value of the position function s(t). The average value of s(t) = t² + 2t on [0,2] is (1/2)∫[0 to 2] (t² + 2t)dt = (1/2)[t³/3 + t²] from 0 to 2 = (1/2)(8/3 + 4) = (1/2)(20/3) = 10/3. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 10/3.
A cyclist’s power output is $P(t)=200-20t$ watts for $0\le t\le 5$. What is the average power?
$\dfrac{P(0)+P(5)}{2}$
$\int_0^5 P(t),dt$
$150$
$P(2.5)$
$\dfrac{1}{5}\int_0^5 P(t),dt$
Explanation
This question asks for the average value of the power function P(t). The average value of P(t) = 200 - 20t on [0,5] is (1/5)∫[0 to 5] (200 - 20t)dt = (1/5)[200t - 10t²] from 0 to 5 = (1/5)(1000 - 250) = 150 watts. Choice C shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 150.
A car’s speed is $v(t)=20+4t$ (mph) for $0\le t\le 3$ hours. What is the average speed?
$v!\left(\dfrac{3}{2}\right)$
$\dfrac{v(0)+v(3)}{2}$
$\int_0^3 v(t),dt$
$26$
$\dfrac{1}{3}\int_0^3 v(t),dt$
Explanation
This question requires using the average value formula for functions. The average value of a function v(t) on interval [a,b] is (1/(b-a))∫[a to b] v(t)dt. For the speed function v(t) = 20 + 4t on [0,3], we compute (1/3)∫[0 to 3] (20 + 4t)dt = (1/3)[20t + 2t²] from 0 to 3 = (1/3)(60 + 18) = 26 mph. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 26.
For $f(x)=e^x$ on $0,1$, which expression equals the average value of $f$ on $0,1$?
$\dfrac{f(0)+f(1)}{2}$
$f!\left(\dfrac{1}{2}\right)$
$\dfrac{1}{1-0}\int_0^1 e^x,dx$
$\dfrac{1}{2}\int_0^1 e^x,dx$
$\int_0^1 e^x,dx$
Explanation
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.
A factory’s output rate is $q(t)=30+6t$ units/hr for $0\le t\le 5$. What is the average output rate?
$\int_0^5 q(t),dt$
$\dfrac{1}{5}\int_0^5 q(t),dt$
$q(2.5)$
$\dfrac{q(0)+q(5)}{2}$
$45$
Explanation
This question asks for the average value of the output rate function q(t). The average value of q(t) = 30 + 6t on [0,5] is (1/5)∫[0 to 5] (30 + 6t)dt = (1/5)[30t + 3t²] from 0 to 5 = (1/5)(150 + 75) = 45 units/hr. Choice C 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 = 45.
A river’s discharge is $R(t)=10+\sin t$ for $0\le t\le 2\pi$. What is the average discharge?
$\int_0^{2\pi} R(t),dt$
$\dfrac{R(0)+R(2\pi)}{2}$
$R(\pi)$
$\dfrac{1}{2\pi}\int_0^{2\pi} R(t),dt$
$10$
Explanation
This question asks for the average value of the discharge function R(t). The average value of R(t) = 10 + sin(t) on [0,2π] is (1/2π)∫[0 to 2π] (10 + sin(t))dt = (1/2π)[10t - cos(t)] from 0 to 2π = (1/2π)(20π + 0) = 10. Since sin(t) integrates to zero over a complete period, only the constant term 10 contributes to the average. Choice D shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 10.
A phone battery level is $B(t)=100-8t$ percent for $0\le t\le 10$. What is the average battery level?
$B(5)$
$\dfrac{B(0)+B(10)}{2}$
$60$
$\int_0^{10} B(t),dt$
$\dfrac{1}{10}\int_0^{10} B(t),dt$
Explanation
This question asks for the average value of the battery level function B(t). The average value of B(t) = 100 - 8t on [0,10] is (1/10)∫[0 to 10] (100 - 8t)dt = (1/10)[100t - 4t²] from 0 to 10 = (1/10)(1000 - 400) = 60 percent. Choice D shows the correct formula setup but doesn't evaluate it. Choice C 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 = 60.
For $f(x)=x^2-4x$ on $0,4$, what is the average value of $f$ on the interval?
$-\dfrac{8}{3}$
$\int_0^4 f(x),dx$
$f(2)$
$\dfrac{1}{4}\int_0^4 f(x),dx$
$\dfrac{f(0)+f(4)}{2}$
Explanation
This problem requires finding the average value of f(x) = x² - 4x on [0,4]. The average value is (1/4)∫[0 to 4] (x² - 4x)dx = (1/4)[x³/3 - 2x²] from 0 to 4 = (1/4)(64/3 - 32) = (1/4)(-32/3) = -8/3. Choice D shows the correct formula setup but doesn't evaluate it. Choice C represents the average of endpoint values, which is not equal to the average value for this nonlinear function. The correct answer is the numerical result E = -8/3.