Calculus 1 Quiz: Average Value Of A Function

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Average Value Of A FunctionQuestion 1 of 20

Find the average value on the interval $[0, 3]$ of the function defined by $f(x) = \begin{cases} x^2 & x < 1 \\ 2x-1 & x \ge 1 \end{cases}$ .

\frac{19}{3}

2

\frac{5}{3}

\frac{19}{9}

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Calculus 1 Quiz

Calculus 1 Quiz: Average Value Of A Function

Practice Average Value Of A Function in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Average Value Of A Function, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

Find the average value on the interval $[0, 3]$ of the function defined by $f(x) = \begin{cases} x^2 & x < 1 \\ 2x-1 & x \ge 1 \end{cases}$ .

$\frac{19}{3}$
$2$
$\frac{5}{3}$
$\frac{19}{9}$ (correct answer)

Explanation: The average value is

\frac{1}{3} \int_0^3 f(x) \,dx

. The integral must be split at

x=1

\int_0^3 f(x) \,dx = \int_0^1 x^2 \,dx + \int_1^3 (2x-1) \,dx

. The first integral is

[\frac{x^3}{3}]_0^1 = \frac{1}{3}

. The second is

[x^2-x]_1^3 = (9-3) - (1-1) = 6

. The total integral is

\frac{1}{3} + 6 = \frac{19}{3}

. The average value is

\frac{1}{3} \cdot \frac{19}{3} = \frac{19}{9}

. Distractor A is the value of the integral. Distractor B results from incorrectly using only the second piece of the function over the whole interval. Distractor C results from incorrectly averaging the average values of the two pieces.

Question 2

Let $A$ be the average value of $f(x) = e^x$ on the interval $[0, 2]$ , and let $M = f(1)$ be the value of the function at the midpoint of the interval. Which of the following statements is true?

$A < M$
$A = M$
$A > M$ (correct answer)
The relationship cannot be determined from the given information.

Explanation: The function

f(x) = e^x

is strictly convex (concave up). For any strictly convex function, the average value over an interval is greater than the function's value at the midpoint of the interval. Therefore,

A > M

. Alternatively, one can calculate the values:

A = \frac{1}{2} \int_0^2 e^x \,dx = \frac{1}{2}[e^x]_0^2 = \frac{e^2 - 1}{2} \approx \frac{7.389-1}{2} \approx 3.19

. The value at the midpoint is

M = e^1 = e \approx 2.718

. Since

3.19 > 2.718

A > M

Question 3

What is the average value of the function $f(x) = x\sqrt{x^2+9}$ on the interval $[0, 4]$ ?

$49/3$
$49/6$ (correct answer)
$98/3$
$4/3$

Explanation: The average value of a function over an interval is a fundamental concept in calculus that requires setting up and evaluating a definite integral. For any function

f(x)

on interval

[a,b]

, the average value equals

\frac{1}{b-a}\int_a^b f(x)\,dx

. For

f(x) = x\sqrt{x^2+9}

[0,4]

, you need to calculate

\frac{1}{4-0}\int_0^4 x\sqrt{x^2+9}\,dx = \frac{1}{4}\int_0^4 x\sqrt{x^2+9}\,dx

. To evaluate this integral, use substitution. Let

u = x^2 + 9

, so

du = 2x\,dx

, which means

x\,dx = \frac{1}{2}du

. When

x = 0

u = 9

; when

x = 4

u = 25

. The integral becomes:

\frac{1}{4} \cdot \frac{1}{2}\int_9^{25} \sqrt{u}\,du = \frac{1}{8}\int_9^{25} u^{1/2}\,du = \frac{1}{8} \cdot \frac{2}{3}u^{3/2}\Big|_9^{25} = \frac{1}{12}[25^{3/2} - 9^{3/2}] = \frac{1}{12}[125 - 27] = \frac{98}{12} = \frac{49}{6}

. This confirms answer B is correct. Choice A (

49/3

) represents doubling the correct answer, likely from forgetting the factor of

\frac{1}{2}

in the substitution. Choice C (

98/3

) omits the

\frac{1}{4}

factor from the average value formula. Choice D (

4/3

) appears to confuse this with a different calculation entirely. Remember: average value problems always require both setting up the integral correctly AND applying the

\frac{1}{b-a}

factor—don't forget either component.

Question 4

If the average value of a continuous function $f(x)$ on the interval $[a, b]$ is $A$ , what is the average value of the function $g(x) = 2f(x) + 3$ on the same interval?

$2A$
$A+3$
$2(A+3)$
$2A+3$ (correct answer)

Explanation: The average value of

g(x)

\frac{1}{b-a} \int_a^b (2f(x) + 3) \,dx

. Using the linearity of integrals, this is

\frac{1}{b-a} [2 \int_a^b f(x) \,dx + \int_a^b 3 \,dx] = 2(\frac{1}{b-a} \int_a^b f(x) \,dx) + \frac{1}{b-a} [3x]_a^b = 2A + \frac{3(b-a)}{b-a} = 2A + 3

Question 5

The velocity of a particle moving along a line is given by $v(t) = 3t^2 - 12$ cm/s for $t \ge 0$ . What is the particle's average velocity from $t=0$ to $t=3$ seconds?

$-9$ cm/s
$-3$ cm/s (correct answer)
$\frac{3}{2}$ cm/s
$\frac{23}{3}$ cm/s

Explanation: Average velocity is the average value of the velocity function.

v_{avg} = \frac{1}{3-0} \int_0^3 (3t^2 - 12) \,dt = \frac{1}{3} [t^3 - 12t]_0^3 = \frac{1}{3}(3^3 - 12(3)) = \frac{1}{3}(27 - 36) = \frac{-9}{3} = -3

cm/s. Distractor A represents the total displacement. Distractor D represents the average speed, which requires integrating

|v(t)|

. Distractor C is the average of the initial and final velocities, a common mistake.

Question 6

For what value of $k > 0$ is the average value of the function $f(x) = \sqrt{x}$ on the interval $[0, k]$ equal to 4?

$6$
$16$
$32$
$36$ (correct answer)

Explanation: The average value is

\frac{1}{k-0} \int_0^k x^{1/2} \,dx = \frac{1}{k} [\frac{2}{3}x^{3/2}]_0^k = \frac{1}{k} (\frac{2}{3}k^{3/2}) = \frac{2}{3}k^{1/2} = \frac{2\sqrt{k}}{3}

. We set this equal to 4:

\frac{2\sqrt{k}}{3} = 4 \implies 2\sqrt{k} = 12 \implies \sqrt{k} = 6 \implies k = 36

. Distractor A is the value of

\sqrt{k}

. Distractor B results from an error in the power rule for integration, using

x^{3/2}

as the antiderivative instead of

\frac{2}{3}x^{3/2}

Question 7

The temperature in degrees Celsius along a 3-meter metal rod, positioned from $x=0$ to $x=3$ , is given by the function $T(x) = 90x(3-x)$ . What is the average temperature of the rod?

$405^{\circ}C$
$202.5^{\circ}C$
$135^{\circ}C$ (correct answer)
$90^{\circ}C$

Explanation: The average temperature is the average value of the function

T(x)

[0, 3]

T_{avg} = \frac{1}{3-0} \int_0^3 (270x - 90x^2) \,dx = \frac{1}{3} [135x^2 - 30x^3]_0^3 = \frac{1}{3} [135(3^2) - 30(3^3)] = \frac{1}{3}[1215 - 810] = \frac{405}{3} = 135

. Distractor A is the value of the definite integral, forgetting to divide by the interval length. Distractor B is the temperature at the midpoint of the rod,

T(1.5)

, which is also the maximum temperature, a common misconception for the average value.

Question 8

For what value of $k > 1$ is the average value of $f(x) = e^x$ on the interval $[0, \ln k]$ equal to $k-1$ ?

$1$
$e$ (correct answer)
$e-1$
$e+1$

Explanation: The average value of

f(x)=e^x

[0, \ln k]

\frac{1}{\ln k - 0} \int_0^{\ln k} e^x \,dx = \frac{1}{\ln k} [e^x]_0^{\ln k} = \frac{e^{\ln k} - e^0}{\ln k} = \frac{k-1}{\ln k}

. We are given that this average value is equal to

k-1

. So,

\frac{k-1}{\ln k} = k-1

. Since

k>1

k-1 \neq 0

, and we can divide both sides by

k-1

to get

\frac{1}{\ln k} = 1

, which implies

\ln k = 1

. Therefore,

k=e

Question 9

The average value of a continuous function $f$ on the interval $[1, 4]$ is $5$ . What is the average value of the function $g(x) = 2f(x) + 3$ on the same interval?

$13$ (correct answer)
$10$
$11$
$39$

Explanation: We are given that the average value of

f

[1, 4]

5

. By definition, this means:

\frac{1}{4-1} \int_1^4 f(x) \, dx = 5

\frac{1}{3} \int_1^4 f(x) \, dx = 5 \implies \int_1^4 f(x) \, dx = 15

We want to find the average value of

g(x) = 2f(x) + 3

[1, 4]

g_{avg} = \frac{1}{4-1} \int_1^4 (2f(x) + 3) \, dx

Using the linearity of integrals:

g_{avg} = \frac{1}{3} \left( \int_1^4 2f(x) \, dx + \int_1^4 3 \, dx \right)

g_{avg} = \frac{1}{3} \left( 2\int_1^4 f(x) \, dx + \int_1^4 3 \, dx \right)

Now, we substitute the known value of the integral of

f(x)

and evaluate the integral of the constant:

\int_1^4 3 \, dx = [3x]_1^4 = 3(4) - 3(1) = 12 - 3 = 9

g_{avg} = \frac{1}{3} (2(15) + 9) = \frac{1}{3} (30 + 9) = \frac{39}{3} = 13

Alternatively, using the property that the average value of

k \cdot f(x) + C

k \cdot f_{avg} + C

, we get

2(5) + 3 = 13

. Distractor B results from calculating

2 \cdot 5

but ignoring the

+3

term. Distractor C results from an error in integrating the constant, calculating

\frac{1}{3}(2(15)+3) = 11

. Distractor D correctly finds the value of the integral

\int_1^4 g(x) dx = 39

but forgets to divide by the interval length of 3.

Question 10

Let $f_{avg}$ be the average value of $f(x)=\sec^2(x)$ on the interval $[0, \pi/4]$ . Let $g_{avg}$ be the average value of $g(x)=\cos(x)$ on the interval $[0, \pi/2]$ . What is the value of the product $f_{avg} \cdot g_{avg}$ ?

$1$
$4/\pi^2$
$8/\pi^2$ (correct answer)
$2/\pi$

Explanation: This question tests your understanding of the average value formula for continuous functions. When finding the average value of a function over an interval, you use the formula: average value =

\frac{1}{b-a}\int_a^b f(x)dx

. For

f_{avg}

, you need the average value of

\sec^2(x)

[0, \pi/4]

. Since

\int \sec^2(x)dx = \tan(x)

, you get:

f_{avg} = \frac{1}{\pi/4 - 0}\int_0^{\pi/4} \sec^2(x)dx = \frac{4}{\pi}[\tan(x)]_0^{\pi/4} = \frac{4}{\pi}(1-0) = \frac{4}{\pi}

For

g_{avg}

, you need the average value of

\cos(x)

[0, \pi/2]

. Since

\int \cos(x)dx = \sin(x)

, you get:

g_{avg} = \frac{1}{\pi/2 - 0}\int_0^{\pi/2} \cos(x)dx = \frac{2}{\pi}[\sin(x)]_0^{\pi/2} = \frac{2}{\pi}(1-0) = \frac{2}{\pi}

Therefore:

f_{avg} \cdot g_{avg} = \frac{4}{\pi} \cdot \frac{2}{\pi} = \frac{8}{\pi^2}

Answer A)

1

likely comes from forgetting to apply the average value formula and just evaluating the functions at specific points. Answer B)

\frac{4}{\pi^2}

results from incorrectly calculating one of the integrals, possibly getting

\tan(\pi/4) = \frac{1}{2}

instead of

1

. Answer D)

\frac{2}{\pi}

is just one of the individual averages, suggesting you forgot to multiply them together. Remember: always set up the average value formula completely before substituting limits, and double-check your antiderivatives and arithmetic when multiplying fractions.

Question 11

According to the Mean Value Theorem for Integrals, there exists a number $c$ in the interval $[1, e]$ such that $f(c)$ equals the average value of the function $f(x) = \frac{1}{x}$ on that interval. What is the value of this number $c$ ?

$\frac{1}{e-1}$
$e-1$ (correct answer)
$\frac{1+e}{2}$
$1$

Explanation: The Mean Value Theorem for Integrals states that if

f

is continuous on

[a, b]

, there exists a

c

[a, b]

such that

f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx

First, calculate the average value of $f(x) = 1/x$ on $[1, e]$ : $f_{avg} = \frac{1}{e-1} \int_1^e \frac{1}{x} \, dx$
The integral of $1/x$ is $\ln|x|$ : $\int_1^e \frac{1}{x} \, dx = [\ln|x|]_1^e = \ln(e) - \ln(1) = 1 - 0 = 1$
Substitute this back into the average value formula: $f_{avg} = \frac{1}{e-1} \cdot 1 = \frac{1}{e-1}$
Now, find the value of $c$ such that $f(c) = f_{avg}$ : $f(c) = \frac{1}{c} = \frac{1}{e-1}$
Solving for $c$ gives: $c = e-1$

Distractor A is the average value of the function,

f_{avg}

, not the value of

c

. Distractor C is the midpoint of the interval, a common but incorrect guess. Distractor D is the value of the definite integral,

\int_1^e f(x) dx

, without dividing by the interval length.

Question 12

The temperature in a room over a 12-hour period is modeled by $T(t) = 68 + 4\sin(\frac{\pi t}{12})$ , where $t$ is in hours from the start time. What is the average temperature, in degrees, during the first 6 hours (from $t=0$ to $t=6$ )?

$70$
$68 - \frac{8}{\pi}$
$68 + \frac{8}{\pi}$ (correct answer)
$68 + \frac{2\pi}{3}$

Explanation: When you encounter a problem asking for average temperature over a time interval, you're dealing with the average value of a function. The average value of a continuous function

f(t)

over interval

[a,b]

is given by

\frac{1}{b-a}\int_a^b f(t)\,dt

. For this problem, you need the average of

T(t) = 68 + 4\sin(\frac{\pi t}{12})

from

t=0

t=6

\text{Average} = \frac{1}{6-0}\int_0^6 \left[68 + 4\sin(\frac{\pi t}{12})\right]dt

= \frac{1}{6}\left[\int_0^6 68\,dt + \int_0^6 4\sin(\frac{\pi t}{12})\,dt\right]

The first integral equals

68 \cdot 6 = 408

. For the second integral, use substitution

u = \frac{\pi t}{12}

, so

du = \frac{\pi}{12}dt

\int_0^6 4\sin(\frac{\pi t}{12})\,dt = 4 \cdot \frac{12}{\pi}\left[-\cos(\frac{\pi t}{12})\right]_0^6 = \frac{48}{\pi}[-\cos(\frac{\pi}{2}) + \cos(0)] = \frac{48}{\pi}[0 + 1] = \frac{48}{\pi}

Therefore:

\text{Average} = \frac{1}{6}\left[408 + \frac{48}{\pi}\right] = 68 + \frac{8}{\pi}

Answer C is correct. Answer A (70) ignores the sine component entirely. Answer B has the wrong sign—it subtracts

\frac{8}{\pi}

instead of adding it. Answer D uses

\frac{2\pi}{3}

instead of

\frac{8}{\pi}

, likely from incorrectly handling the integral bounds or substitution. Study tip: When finding average values involving trigonometric functions, always check your integration limits carefully after substitution, and remember that the average value formula requires dividing by the interval length.

Question 13

Let $f(x) = \sqrt{x}$ . Let $A_1$ be the average value of $f$ on the interval $[0, 1]$ and let $A_2$ be the average value of $f$ on the interval $[1, 4]$ . Which of the following statements correctly describes the relationship between $A_1$ and $A_2$ ?

$A_1 > A_2$
$A_1 < A_2$ (correct answer)
$A_1 = A_2$
The relationship cannot be determined without computation.

Explanation: The function

f(x) = \sqrt{x}

is a strictly increasing function for

x > 0

. The average value of a strictly increasing function over an interval

[c,d]

will always be greater than its average value over a preceding interval

[a,b]

(where

b \le c

). Therefore,

A_1 < A_2

. By computation:

A_1 = \frac{1}{1} \int_0^1 x^{1/2} \,dx = [\frac{2}{3}x^{3/2}]_0^1 = \frac{2}{3}

A_2 = \frac{1}{3} \int_1^4 x^{1/2} \,dx = \frac{1}{3}[\frac{2}{3}x^{3/2}]_1^4 = \frac{2}{9}(8-1) = \frac{14}{9}

. Since

\frac{2}{3} = \frac{6}{9}

, we have

\frac{6}{9} < \frac{14}{9}

, confirming that

A_1 < A_2

Question 14

The area of the region under the curve of a positive continuous function $y=f(x)$ from $x=1$ to $x=9$ is 32 square units. What is the height $h$ of a rectangle with a base on the interval $[1, 9]$ that has the same area?

$4$ (correct answer)
$8$
$32$
$256$

Explanation: The area under the curve is given by the definite integral

\int_1^9 f(x) \,dx = 32

. A rectangle with the same area over the same base

[1, 9]

would have area

A = h \times (9-1) = 8h

. Setting the areas equal,

8h = 32

, which means

h=4

. This height

h

is precisely the average value of the function

f(x)

on the interval

[1, 9]

Question 15

The average value of the derivative of a differentiable function, $f'(x)$ , on an interval $[a, b]$ represents which quantity?

The average value of the original function $f(x)$ on $[a, b]$ .
The total change in the function, $f(b) - f(a)$ .
The average rate of change of the function $f(x)$ on $[a, b]$ . (correct answer)
The instantaneous rate of change at the midpoint, $f'(\frac{a+b}{2})$ .

Explanation: The average value of

f'(x)

[a, b]

is given by the expression

\frac{1}{b-a} \int_a^b f'(x) \,dx

. By the Fundamental Theorem of Calculus,

\int_a^b f'(x) \,dx = f(b) - f(a)

. Therefore, the expression becomes

\frac{f(b)-f(a)}{b-a}

, which is the definition of the average rate of change of

f(x)

on the interval

[a, b]

Question 16

What is the average value of the function $f(x) = x^3 + 2x^2 - x$ on the interval $[-2, 2]$ ?

$0$
$\frac{11}{6}$
$\frac{8}{3}$ (correct answer)
$\frac{32}{3}$

Explanation: The average value is

\frac{1}{2 - (-2)} \int_{-2}^2 (x^3 + 2x^2 - x) \,dx = \frac{1}{4} \int_{-2}^2 (x^3 + 2x^2 - x) \,dx

. Since the integral is over a symmetric interval

[-a, a]

, the integral of the odd parts of the function (

x^3

and

-x

) is zero. We only need to integrate the even part:

\frac{1}{4} \int_{-2}^2 2x^2 \,dx = \frac{1}{2} \int_{-2}^2 x^2 \,dx = \frac{1}{2} [\frac{x^3}{3}]_{-2}^2 = \frac{1}{2} (\frac{8}{3} - \frac{-8}{3}) = \frac{1}{2}(\frac{16}{3}) = \frac{8}{3}

. Distractor A results from incorrectly assuming the entire function is odd. Distractor D is the value of the integral without dividing by the interval length. Distractor B results from integrating over

[0,2]

instead of

[-2,2]

Question 17

Suppose $\int_{-2}^{2} f(x) \,dx = 8$ and $\int_{-2}^{2} g(x) \,dx = -10$ . Find the average value of $h(x) = 3f(x) - \frac{1}{2}g(x)$ on the interval $[-2, 2]$ .

$\frac{19}{4}$
$\frac{29}{2}$
$29$
$\frac{29}{4}$ (correct answer)

Explanation: The interval is

[-2, 2]

, so its length is

2 - (-2) = 4

. The average value is

\frac{1}{4} \int_{-2}^2 (3f(x) - \frac{1}{2}g(x)) \,dx

. Using linearity of integrals, this is

\frac{1}{4} [3 \int_{-2}^2 f(x) \,dx - \frac{1}{2} \int_{-2}^2 g(x) \,dx] = \frac{1}{4} [3(8) - \frac{1}{2}(-10)] = \frac{1}{4}[24 + 5] = \frac{29}{4}

. Distractor C is the value of the integral of

h(x)

without dividing by the interval length. Distractor B divides by an incorrect interval length of 2. Distractor A results from a sign error in the calculation.

Question 18

The average value of the function $f(x) = kx^2 + 1$ on the interval $[0, 3]$ is $13$ . What is the value of the constant $k$ ?

$4$ (correct answer)
$10/9$
$13/3$
$14/3$

Explanation: The average value of a function

f(x)

on an interval

[a, b]

is given by the formula

f_{avg} = \frac{1}{b-a} \int_a^b f(x) \, dx

. We are given that the average value is

13

on the interval

[0, 3]

Set up the equation based on the formula: $13 = \frac{1}{3-0} \int_0^3 (kx^2 + 1) \, dx$
Calculate the definite integral: $\int_0^3 (kx^2 + 1) \, dx = \left[ \frac{k}{3}x^3 + x \right]_0^3 = \left( \frac{k}{3}(3)^3 + 3 \right) - (0) = 9k + 3$
Substitute the result back into the average value equation and solve for $k$ : $13 = \frac{1}{3}(9k + 3)$ $13 = 3k + 1$ $12 = 3k$ $k = 4$

Distractor B arises from forgetting to divide by the interval length

(b-a)=3

, leading to the incorrect equation

13 = 9k+3

. Distractor C results from ignoring the '

+1

' term in the function

f(x)

. Distractor D is the result of a simple arithmetic error when solving

13 = 3k+1

Question 19

The velocity of a particle moving along a straight line is given by $v(t) = 3t^2 - 12$ for time $t \ge 0$ . What is the average speed of the particle over the interval $[0, 3]$ ?

$23/3$ (correct answer)
$-3$
$23$
$27/2$

Explanation: Average speed is the average value of the speed function, which is the absolute value of velocity,

|v(t)|

. The formula is

\frac{1}{b-a} \int_a^b |v(t)| \, dt

First, determine where the velocity $v(t) = 3t^2 - 12$ is negative on the interval $[0, 3]$ . Set $v(t) = 0$ : $3t^2 - 12 = 0 \implies t^2 = 4 \implies t=2$ (since $t \ge 0$ ). The function $v(t)$ is negative on $[0, 2)$ and positive on $(2, 3]$ .
Set up the integral for average speed, splitting it at $t=2$ to handle the absolute value: $\text{Average Speed} = \frac{1}{3-0} \int_0^3 |3t^2 - 12| \, dt = \frac{1}{3} \left( \int_0^2 -(3t^2 - 12) \, dt + \int_2^3 (3t^2 - 12) \, dt \right)$
Evaluate the integrals: $\int_0^2 (12 - 3t^2) \, dt = [12t - t^3]_0^2 = (24 - 8) - 0 = 16$ $\int_2^3 (3t^2 - 12) \, dt = [t^3 - 12t]_2^3 = (27 - 36) - (8 - 24) = -9 - (-16) = 7$
Sum the results and find the average: $\text{Average Speed} = \frac{1}{3} (16 + 7) = \frac{23}{3}$

Distractor B calculates the average velocity, not average speed, which is

\frac{1}{3}\int_0^3 (3t^2 - 12) dt = -3

. Distractor C correctly calculates the total distance traveled,

\int_0^3 |v(t)| dt = 23

, but forgets to divide by the interval length. Distractor D is the average of the initial and final speeds,

\frac{|v(0)|+|v(3)|}{2}

, a common misconception.

Question 20

Let $f(x) = 3x^2 - 2x$ . The average value of $f$ on the interval $[0, k]$ is equal to the value of the function at $x=k$ , where $k>0$ . What is the value of $k$ ?

$2 - \sqrt{2}$
$1$
$2/3$
$1/2$ (correct answer)

Explanation: This problem combines the average value of a function with the Mean Value Theorem concept. When you see "average value equals function value," you're looking for a point where the function's actual value matches its average over an interval. The average value of

f(x) = 3x^2 - 2x

[0, k]

is:

\text{Average} = \frac{1}{k-0} \int_0^k (3x^2 - 2x) \, dx = \frac{1}{k} \left[ x^3 - x^2 \right]_0^k = \frac{k^3 - k^2}{k} = k^2 - k

We need this average to equal

f(k) = 3k^2 - 2k

k^2 - k = 3k^2 - 2k

k^2 - k - 3k^2 + 2k = 0

-2k^2 + k = 0

k(-2k + 1) = 0

Since

k > 0

, we have

-2k + 1 = 0

, so

k = \frac{1}{2}

. This is choice (D). Choice (A)

2 - \sqrt{2}

might tempt you if you incorrectly set up a quadratic formula problem or made algebraic errors. Choice (B)

1

could result from solving

k^2 - k = 0

and choosing the wrong root, forgetting that we need the equation

k^2 - k = 3k^2 - 2k

. Choice (C)

\frac{2}{3}

might come from inverting the correct answer or making sign errors in the algebraic manipulation. Remember: average value problems often lead to polynomial equations. Always set up the integral correctly, simplify the average value expression completely, then set it equal to the function value at the endpoint. Double-check your algebra—these problems are more about careful computation than complex concepts.