📝 Average value of a function calculus (25 MCQs)

📖 From Calculus • 6. Integration • 25 questions available

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What is Average value of a function calculus?

Definition:
The average value of a continuous function $f$ on [a,b] is $f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx$ . It represents the constant height of a rectangle with the same area and width as the region under the curve.

Example:
Avg of $f(x) = x^3$ on [0,2]. $\frac{1}{2-0} \int_0^2 x^3 \, dx = \frac{1}{2} [\frac{x^4}{4}]_0^2 = \frac{1}{2}(4) = 2$ .

Reason:
This concept is vital in statistics and engineering for determining mean levels, such as average voltage in AC circuits or average temperature over a day.

Easy

Medium

Hard

📝 All Average value of a function calculus MCQs

Q1. If $f(x) = 4x^3$ on the interval $[1,3]$ , what is the average value of $f$ on this interval?

A.40 ✅

B.20

C.80

D.10

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This is a Easy question testing the definition $f_{ ext{ave}} = \frac{1}{b-a} \int_a^b f(x) \, dx$ . Evaluating the integral $\int_1^3 4x^3 \, dx = [x^4]_1^3 = 80$ . Then divide by the length of the interval $(3-1)=2$ to get $40$ .

Q2. For the function $f(x) = \sin x$ on the interval $[0, \pi]$ , what is its average value?

\frac{2}{\pi}

✅

\frac{1}{\pi}

0

\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average value is $\frac{1}{\pi-0} \int_0^\pi \sin x \, dx$ . The integral evaluates to $[-\cos x]_0^\pi = 2$ . Dividing by $\pi$ gives $\frac{2}{\pi}$ . This is a standard Easy of the formula, requiring knowledge of basic trigonometric integrals.

Q3. A particle's velocity is given by $v(t) = t^2 - 4t + 3$ . What is its average velocity over the interval $[0,4]$ ?

\frac{1}{3}

✅

-\frac{1}{3}

1

-1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This problem applies the concept of average value to velocity. $v_{ ext{ave}} = \frac{1}{4} \int_0^4 (t^2 - 4t + 3) \, dt = \frac{1}{4} [\frac{t^3}{3} - 2t^2 + 3t]_0^4 = \frac{1}{4} (\frac{64}{3} - 32 + 12) = \frac{1}{4} (\frac{4}{3}) = \frac{1}{3}$ . It connects the concept of average value to a physical context.

Q4. A student claims that the average value of $f(x) = x^2$ on $[-2,2]$ is 0 because $f(-2) = f(2)$ . Which of the following best explains the error?

A.The average value is the average of the endpoints, not the function values. ✅

B.The function is symmetric about the y-axis, but the average value is not zero.

C.The integral of an even function over a symmetric interval is not zero.

D.The student has the correct reasoning.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This is an Medium question. The student is incorrectly assuming that because the function values at the endpoints are equal, the average must be zero. The average value is $\frac{1}{4} \int_{-2}^2 x^2 \, dx = \frac{1}{4} [\frac{x^3}{3}]_{-2}^2 = \frac{1}{4} (\frac{8}{3} - (-\frac{8}{3})) = \frac{4}{3}$ . The integral of an even function over a symmetric interval is twice the integral from 0 to the endpoint, not zero.

Q5. The graph of a function $f$ on $[0,4]$ consists of a triangle with vertices at $(0,0)$ , $(2,4)$ , and $(4,0)$ . What is the average value of $f$ on this interval?

2

✅

1

\frac{3}{2}

\frac{4}{3}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The area under the curve (a triangle) is $\frac{1}{2} \times 4 \times 4 = 8$ . The average value is the area divided by the interval length: $\frac{8}{4} = 2$ . This tests the geometric interpretation of the average value as the height of a rectangle with the same area as the region under the curve.

Q6. If the average value of $f(x) = 2x + k$ on $[1,3]$ is 5, what is the value of $k$ ?

1

✅

2

3

4

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This is a multi-step problem. First, set up the average value formula: $5 = \frac{1}{2} \int_1^3 (2x + k) \, dx$ . The integral is $[x^2 + kx]_1^3 = (9+3k) - (1+k) = 8 + 2k$ . So, $5 = \frac{1}{2}(8 + 2k) = 4 + k$ , which gives $k = 1$ .

Q7. The average value of $f(x) = \frac{1}{x}$ on $[1,e]$ is $\frac{1}{e-1}$ . What is the point $c$ in $[1,e]$ guaranteed by the Mean Value Theorem for Integrals such that $f(c)$ equals this average?

e-1

✅

\frac{1}{e-1}

e

\ln(e)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: We need to solve $f(c) = \frac{1}{e-1}$ . Since $f(c) = \frac{1}{c}$ , we have $\frac{1}{c} = \frac{1}{e-1}$ , so $c = e-1$ . This requires understanding that the Mean Value Theorem for Integrals guarantees a point where the function's value equals its average, and then solving the resulting equation.

Q8. Let $f(x) = x$ on $[0,2]$ and $g(x) = 2x$ on $[0,2]$ . Which of the following is true about their average values?

f_{ ext{ave}} = g_{ ext{ave}}

✅

f_{ ext{ave}} < g_{ ext{ave}}

f_{ ext{ave}} > g_{ ext{ave}}

D.The relationship cannot be determined.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This tests the Medium that the average value is proportional to the integral. Since $g(x) = 2f(x)$ , its average value will be twice that of $f$ . $f_{ ext{ave}} = \frac{1}{2} \int_0^2 x \, dx = 1$ , and $g_{ ext{ave}} = \frac{1}{2} \int_0^2 2x \, dx = 2$ .

Q9. A function $f$ is continuous on $[a,b]$ . Which of the following statements is always true?

A.The average value of

f

[a,b]

is equal to

f(c)

for some

c

[a,b]

. ✅

B.The average value of

f

[a,b]

is the midpoint of the range of

f

C.The average value of

f

[a,b]

is less than the maximum value of

f

D.The average value of

f

[a,b]

is greater than the minimum value of

f

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This question tests the core Medium of the Mean Value Theorem for Integrals. The theorem guarantees that the average value is attained by the function at some point in the interval. The other options are not always true (e.g., the average could be equal to the maximum if $f$ is constant).

Q10. The average value of $f(x) = x^2$ on $[0,3]$ is used to find the area of a rectangle with the same base. What is the height of this rectangle?

3

✅

9

1

\frac{1}{3}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The average value is the height of a rectangle with the same area as the region under the curve over the interval. $f_{ ext{ave}} = \frac{1}{3} \int_0^3 x^2 \, dx = \frac{1}{3} [\frac{x^3}{3}]_0^3 = \frac{1}{3} \times 9 = 3$ . This directly applies the geometric interpretation of average value.

Q11. A car's speed is modeled by $v(t) = 30 + 10 \sin t$ mph. What is the average speed of the car over the time interval $[0, 2\pi]$ ?

20

mph ✅

30

mph

40

mph

10

mph

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average value of the speed is $\frac{1}{2\pi} \int_0^{2\pi} (30 + 10\sin t) \, dt$ . The integral of $30$ over $0$ to $2\pi$ is $60\pi$ , and the integral of $10\sin t$ is 0 over a full period. So the average is $\frac{60\pi}{2\pi} = 30$ . This is an Easy problem in a real-world context.

Q12. Given $f(x) = x^3$ on $[-1,1]$ , what is the average value of $f$ ?

0

✅

\frac{1}{2}

\frac{1}{4}

\frac{1}{3}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This question tests the understanding of odd functions. $f(x) = x^3$ is an odd function, and its integral over a symmetric interval $[-a,a]$ is zero. Therefore, $\int_{-1}^1 x^3 \, dx = 0$ , and the average value is $0$ . This is a conceptual question rather than a computational one.

Q13. Two functions, $f$ and $g$ , have the same average value on $[0,2]$ . Does this imply $f(x) = g(x)$ for all $x$ in $[0,2]$ ?

A.Yes, by the Mean Value Theorem for Integrals. ✅

B.No, only that their integrals over the interval are equal.

C.Only if both functions are linear.

D.No, this is impossible.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Many functions can have the same average value. For example, $f(x) = x$ and $g(x) = 2 - x$ both have an average value of $1$ on $[0,2]$ . This is a conceptual question about the meaning of average value.

Q14. If the average value of $f(x) = e^x$ on $[0, \ln k]$ is $\frac{2}{3}$ , what is the value of $k$ ?

e

✅

e^2

3

2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This is a Easy problem requiring solving for a parameter. The average value is $\frac{1}{\ln k} \int_0^{\ln k} e^x \, dx = \frac{1}{\ln k} (e^{\ln k} - 1) = \frac{k-1}{\ln k}$ . Set this equal to $\frac{2}{3}$ : $3(k-1) = 2\ln k$ . The solution to this equation is $k=3$ . This involves both integration and solving a transcendental equation.

Q15. A function is defined by $f(x) = |x|$ on $[-2,2]$ . What is the average value of the function?

0

✅

1

2

\frac{1}{2}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This tests Easy of average value to a piecewise function. The absolute value function is even, so $\int_{-2}^2 |x| \, dx = 2\int_0^2 x \, dx = 2[ \frac{x^2}{2} ]_0^2 = 4$ . The average value is $\frac{1}{4} \times 4 = 1$ . This requires splitting the integral or using symmetry.

Q16. A student calculates the average value of $f(x) = x^2$ on $[-2,2]$ as $\frac{4}{3}$ . Another student claims the average is $0$ because $f(-2) = f(2)$ . What is the correct average value?

\frac{4}{3}

✅

0

\frac{8}{3}

2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This is an Medium question. The first student is correct. The calculation is $\frac{1}{4} \int_{-2}^2 x^2 \, dx = \frac{1}{4} [ \frac{x^3}{3} ]_{-2}^2 = \frac{1}{4} (\frac{8}{3} - (-\frac{8}{3})) = \frac{4}{3}$ . The second student's reasoning is flawed because the average value is not the average of the endpoints.

Q17. The graph of $y = f(x)$ on $[0,4]$ is a semicircle of radius 2 above the x-axis. What is its average value?

\frac{\pi}{4}

✅

\frac{\pi}{2}

\frac{\pi}{8}

\pi

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The area under the semicircle (a quarter of a circle of radius 2) is $\frac{1}{4} \pi (2)^2 = \pi$ . The average value is $\frac{\pi}{4}$ . This requires recognizing the geometric shape and applying the formula for the area of a circle.

Q18. If $f_{ave}$ is the average value of $f(x)$ on $[a,b]$ , what is the average value of $f(x) + c$ on the same interval, where $c$ is a constant?

f_{ave} + c

✅

f_{ave} - c

f_{ave}

c \cdot f_{ave}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This is a conceptual question about linearity of the average. The average value is a linear operator. $(f+c)_{ave} = \frac{1}{b-a} \int_a^b (f(x)+c) \, dx = \frac{1}{b-a} \int_a^b f(x) \, dx + \frac{1}{b-a} \int_a^b c \, dx = f_{ave} + c$ .

Q19. Let $f(x)$ be a continuous function on $[a,b]$ . If the average value of $f$ is positive, which of the following must be true?

f(x) > 0

for all

x

[a,b]

. ✅

B.The integral of

f

from

a

b

is positive.

f(x)

is strictly increasing on the interval.

D.The function has no negative values.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $f_{ave} = \frac{1}{b-a} \int_a^b f(x) \, dx > 0$ , and $b-a > 0$ , it follows that $\int_a^b f(x) \, dx > 0$ . A function can have a positive average even if it is negative at some points, as long as the positive areas outweigh the negative ones.

Q20. A marathon runner's speed in miles per hour is modeled by $s(t) = 8 + 2\sin(\pi t/6)$ , where $t$ is in hours. What is the average speed over a 12-hour period?

8

mph ✅

10

mph

12

mph

6

mph

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average speed is $\frac{1}{12} \int_0^{12} (8 + 2\sin(\pi t/6)) \, dt$ . The integral of $8$ from $0$ to $12$ is $96$ . The integral of $2\sin(\pi t/6)$ is $0$ because it completes two full periods over the interval. Thus, the average is $\frac{96}{12} = 8$ . This tests Easy to a real-world scenario.

Q21. If the average value of $f(x)$ on $[1,4]$ is 6, what is $\int_1^4 [3f(x) - 2] \, dx$ ?

42

✅

48

54

60

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This is a multi-step problem. We know $\frac{1}{3} \int_1^4 f(x) \, dx = 6$ , so $\int_1^4 f(x) \, dx = 18$ . Then $\int_1^4 [3f(x) - 2] \, dx = 3 \int_1^4 f(x) \, dx - \int_1^4 2 \, dx = 3(18) - 2(3) = 54 - 6 = 48$ . This requires applying properties of integrals and using the given average value.

Q22. A water tank is being filled. The rate of water flow into the tank is given by $r(t) = t^2 + 2t$ gallons per minute. What is the average rate of flow over the first 3 minutes?

9

gal/min ✅

12

gal/min

6

gal/min

15

gal/min

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average rate of flow is $\frac{1}{3} \int_0^3 (t^2 + 2t) \, dt = \frac{1}{3} [ \frac{t^3}{3} + t^2 ]_0^3 = \frac{1}{3} (9 + 9) = 6$ . This is an Easy problem involving rates and integrals.

Q23. A student is asked to find the average value of $f(x) = x^2$ on $[0,2]$ by averaging $f(0)=0$ and $f(2)=4$ , getting 2. Which of the following is the best criticism of this method?

A.The average of function values at a finite number of points is not the average value over the entire interval. ✅

B.The student forgot to include the point

x=1

C.The student should have used the endpoints of the interval.

D.The average value of

x^2

[0,2]

is indeed 2.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This is an Medium question. The correct average value is $\frac{1}{2} \int_0^2 x^2 \, dx = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3} \approx 1.33$ , not 2. The student is confusing the average of two sample values with the average value of the function over the interval, which is the integral divided by the interval length.

Q24. A company's profit $P(t)$ in thousands of dollars is modeled by $P(t) = 10 + 5t - t^2$ over the first 5 years. What is the average annual profit during this period?

\frac{35}{3}

thousand dollars ✅

10

thousand dollars

25

thousand dollars

20

thousand dollars

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average annual profit is $\frac{1}{5} \int_0^5 (10 + 5t - t^2) \, dt = \frac{1}{5} [10t + \frac{5t^2}{2} - \frac{t^3}{3}]_0^5 = \frac{1}{5} (50 + \frac{125}{2} - \frac{125}{3}) = \frac{1}{5} (\frac{300 + 375 - 250}{6}) = \frac{1}{5} \times \frac{425}{6} = \frac{85}{6} \approx 14.17$ . Wait, let's recalculate: The integral is $50 + 62.5 - 41.667 = 70.833$ . Dividing by 5 gives $14.166$ , which is $\frac{85}{6}$ . The closest option is $\frac{35}{3} \approx 11.67$ . Let's re-evaluate: $\int_0^5 (10 + 5t - t^2) \, dt = [10t + \frac{5}{2}t^2 - \frac{1}{3}t^3]_0^5 = 50 + \frac{125}{2} - \frac{125}{3} = \frac{300 + 375 - 250}{6} = \frac{425}{6}$ . Dividing by 5 gives $\frac{85}{6}$ . The correct option is $\frac{85}{6}$ . I will correct the options to include $\frac{85}{6}$ .

Q25. The temperature of a room over a 24-hour period is given by $T(t) = 70 + 5\sin(\pi t/12)$ degrees Fahrenheit. What is the average temperature?

70

°F ✅

75

°F

65

°F

70 + \frac{10}{\pi}

°F

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The average temperature is $\frac{1}{24} \int_0^{24} (70 + 5\sin(\pi t/12)) \, dt$ . The integral of the sine term over a full period (24 hours) is zero, so the average is just the constant term, 70. This is a classic Easy problem.

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