AP Calculus BC Practice Quiz: Finding the Average Value of a Function on an Interval
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) \int_{1}^{5} x^3 dx
B) \frac{1}{5} \int_{1}^{5} x^3 dx
C) \frac{1}{4} \int_{1}^{5} x^3 dx
D) 4 \int_{1}^{5} x^3 dx
Correct Answer: C
The average value of a continuous function $f$ over an interval $[a, b]$ is given by the formula $\frac{1}{b-a} \int_{a}^{b} f(x) dx$. For the function $f(x) = x^3$ on the interval $[1, 5]$, we have $a=1$ and $b=5$. Therefore, $b-a = 5-1 = 4$. The correct expression is $\frac{1}{4} \int_{1}^{5} x^3 dx$.
A) 4
B) 5
C) 10
D) 20
Correct Answer: B
To find the average value, we use the formula $\frac{1}{b-a} \int_{a}^{b} f(x) dx$. Here, $a=0$, $b=4$, and $f(x) = 2x+1$. The setup is $\frac{1}{4-0} \int_{0}^{4} (2x+1) dx$. The integral of $2x+1$ is $x^2+x$. Evaluating from 0 to 4 gives $[x^2+x]_{0}^{4} = (4^2+4) - (0^2+0) = 16+4 = 20$. The average value is $\frac{1}{4}(20) = 5$.
A) 0
B) 2
C) \frac{2}{\pi}
D) \frac{\pi}{2}
Correct Answer: C
The average value is $\frac{1}{\pi-0} \int_{0}^{\pi} \sin(x) dx$. The antiderivative of $\sin(x)$ is $-\cos(x)$. Evaluating the definite integral: $[-\cos(x)]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1+1=2$. Multiplying by $\frac{1}{\pi}$ gives the average value of $\frac{2}{\pi}$.
A) \frac{1}{e^2-1}
B) \frac{2}{e^2-1}
C) \frac{1}{e^2}
D) 2
Correct Answer: B
Using the average value formula, we calculate $\frac{1}{e^2-1} \int_{1}^{e^2} \frac{1}{x} dx$. The integral of $\frac{1}{x}$ is $\ln|x|$. Evaluating from 1 to $e^2$: $[\ln|x|]_{1}^{e^2} = \ln(e^2) - \ln(1) = 2 - 0 = 2$. Therefore, the average value is $\frac{1}{e^2-1} \cdot 2 = \frac{2}{e^2-1}$.
A) [-3, 4]
B) [0, 7]
C) [-3, 7]
D) [4, 7]
Correct Answer: A
The formula for the average value of a function $f$ on an interval $[a, b]$ is $\frac{1}{b-a} \int_{a}^{b} f(t) dt$. By comparing this to the given expression, we can identify the limits of integration as $a=-3$ and $b=4$. The coefficient $\frac{1}{b-a}$ should be $\frac{1}{4 - (-3)} = \frac{1}{7}$. Since this matches the given expression, the interval is $[-3, 4]$.
A) -1
B) 3
C) -3
D) 1
Correct Answer: A
The average value is given by $\frac{1}{2 - (-1)} \int_{-1}^{2} (3x^2 - 4) dx = \frac{1}{3} \int_{-1}^{2} (3x^2 - 4) dx$. The antiderivative is $x^3 - 4x$. Evaluating the definite integral: $[x^3 - 4x]_{-1}^{2} = (2^3 - 4(2)) - ((-1)^3 - 4(-1)) = (8 - 8) - (-1 + 4) = 0 - 3 = -3$. The average value is $\frac{1}{3}(-3) = -1$.
A) \frac{5}{\ln(5)}
B) \frac{4}{\ln(5)}
C) 4
D) 5
Correct Answer: B
The average value is calculated using the formula $\frac{1}{b-a} \int_{a}^{b} h(x) dx$. Here, $a=0$ and $b=\ln(5)$. The setup is $\frac{1}{\ln(5) - 0} \int_{0}^{\ln(5)} e^x dx$. The integral of $e^x$ is $e^x$. Evaluating the definite integral: $[e^x]_{0}^{\ln(5)} = e^{\ln(5)} - e^0 = 5 - 1 = 4$. Therefore, the average value is $\frac{1}{\ln(5)} \cdot 4 = \frac{4}{\ln(5)}$.