Average Value of Functions on Intervals
Help Questions
AP Calculus AB › Average Value of Functions on Intervals
For $f(x)=\dfrac{1}{2}x$ on $4,8$, what is the average value of $f$ on the interval?
$3$
$\dfrac{f(4)+f(8)}{2}$
$\int_4^8 f(x),dx$
$f(6)$
$\dfrac{1}{4}\int_4^8 f(x),dx$
Explanation
This problem requires finding the average value of f(x) = x/2 on [4,8]. The average value is (1/4)∫[4 to 8] (x/2)dx = (1/4)[(x²/4)] from 4 to 8 = (1/4)(16 - 4) = 3. 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 = 3.
A tank’s volume is $V(t)=100+5t- t^2$ for $0\le t\le 5$. What is the average value of $V$?
$\dfrac{1}{5}\int_0^5 V(t),dt$
$V(2.5)$
$\dfrac{V(0)+V(5)}{2}$
$\int_0^5 V(t),dt$
$\dfrac{335}{6}$
Explanation
This question asks for the average value of the volume function V(t). The average value of V(t) = 100 + 5t - t² on [0,5] is (1/5)∫[0 to 5] (100 + 5t - t²)dt = (1/5)[100t + (5t²)/2 - t³/3] from 0 to 5 = (1/5)(500 + 62.5 - 125/3) = (1/5)(1687.5/3) = 335/6. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 335/6.
A river’s discharge is $R(t)=10+\sin t$ for $0\le t\le 2\pi$. What is the average discharge?
$\dfrac{R(0)+R(2\pi)}{2}$
$\dfrac{1}{2\pi}\int_0^{2\pi} R(t),dt$
$R(\pi)$
$10$
$\int_0^{2\pi} R(t),dt$
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.
If $g(x)=x^2$ on $0,3$, what is the average value of $g$ on $0,3$?
$3$
$\int_0^3 g(x),dx$
$\dfrac{g(0)+g(3)}{2}$
$g(1.5)$
$\dfrac{1}{3}\int_0^3 g(x),dx$
Explanation
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.
A population model is $P(t)=1000(1+t)$ for $0 \le t \le 2$. What is the average population on $0,2$?
$\dfrac{P(0)+P(2)}{2}$
$\dfrac{1}{2}\int_0^2 P(t),dt$
$2000$
$\int_0^2 P(t),dt$
$P(1)$
Explanation
This question asks for the average value of the population function P(t). The average value of $P(t) = 1000(1 + t)$ on [0,2] is $\frac{1}{2} \int_0^2 1000(1 + t) , dt = \frac{1}{2} [1000t + 500t^2]_0^2 = \frac{1}{2}(2000 + 2000) = 2000$. 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 = 2000.
For $p(x)=\dfrac{1}{1+x^2}$ on $0,1$, which expression gives the average value of $p$?
$\dfrac{1}{2}\int_0^1 p(x),dx$
$\int_0^1 p(x),dx$
$p!\left(\dfrac{1}{2}\right)$
$\dfrac{1}{1-0}\int_0^1 p(x),dx$
$\dfrac{p(0)+p(1)}{2}$
Explanation
This problem asks for the correct expression representing the average value of p(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] p(x)dx = ∫[0 to 1] p(x)dx. Choice E incorrectly includes an extra factor of 1/2. The correct expression is choice D, which properly applies the average value formula.
For $f(x)=3x+1$ on $2,6$, what is the average value of $f$ on the interval?
$13$
$f(4)$
$\dfrac{1}{4}\int_2^6 f(x),dx$
$\int_2^6 f(x),dx$
$\dfrac{f(2)+f(6)}{2}$
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 $0,2$?
$\dfrac{10}{3}$
$\dfrac{s(0)+s(2)}{2}$
$s(1)$
$\int_0^2 s(t),dt$
$\dfrac{1}{2}\int_0^2 s(t),dt$
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 machine’s cost rate is $c(t)=12+3t$ dollars/hour for $2\le t\le 6$. What is the average cost rate?
$c(4)$
$\int_2^6 c(t),dt$
$\dfrac{1}{4}\int_2^6 c(t),dt$
$\dfrac{c(2)+c(6)}{2}$
$24$
Explanation
This question asks for the average value of the cost rate function $c(t)$. The average value of $c(t) = 12 + 3t$ on $[2,6]$ is $\frac{1}{4} \int_2^6 (12 + 3t) , dt$ = $\frac{1}{4} [12t + \frac{3t^2}{2}] \big|_{2}^{6}$ = $\frac{1}{4} [(72 + 54) - (24 + 6)]$ = $\frac{1}{4} (96)$ = 24 dollars/hour. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 24.
For $f(x)=\dfrac{2}{x}$ on $1,4$, which expression represents the average value of $f$ on $1,4$?
$\dfrac{f(1)+f(4)}{2}$
$f(2.5)$
$\dfrac{1}{4}\int_1^4 \dfrac{2}{x},dx$
$\int_1^4 \dfrac{2}{x},dx$
$\dfrac{1}{3}\int_1^4 \dfrac{2}{x},dx$
Explanation
This problem asks for the correct expression representing the average value of f(x) = 2/x on [1,4]. 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 [1,4], this becomes (1/(4-1))∫[1 to 4] (2/x)dx = (1/3)∫[1 to 4] (2/x)dx. Choice E incorrectly uses 1/4 as the coefficient instead of 1/3. The correct expression is choice A, which properly applies the average value formula.