Average Value of a Function - Calculus 1

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 $A_2$ ?
Answer: $A_1 < A_2$ 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 $A_1 < A_2$ .
The average value of the function $f(x) = k\sqrt{x}$ on the interval $[1, 4]$ is 7. What is the value of the constant $k$ ?
Answer: $\frac{9}{2}$ The average value is given by $f_{avg} = \frac{1}{4-1} \int_1^4 k\sqrt{x} \,dx$ . We set this equal to 7. $7 = \frac{k}{3} \int_1^4 x^{1/2} \,dx = \frac{k}{3} [\frac{2}{3}x^{3/2}]_1^4 = \frac{2k}{9}[4^{3/2} - 1^{3/2}] = \frac{2k}{9}[8-1] = \frac{14k}{9}$ . Solving for $k$ : $7 = \frac{14k}{9} \implies 63 = 14k \implies k = \frac{63}{14} = \frac{9}{2}$ . Distractor A results from forgetting the $\frac{1}{b-a}$ factor. Distractor B results from using an incorrect interval length of 4. Distractor C results from an error in evaluating the exponent.
For what value of $k > 1$ is the average value of $f(x) = e^x$ on the interval $[0, \ln k]$ equal to $k-1$ ?
Answer: $e$ The average value of $f(x)=e^x$ on $[0, \ln k]$ is $\frac{1}{\ln k - 0} \int_0^{\ln k} e^x \,dx = \frac{1}{\ln k} [e^x]_0^{\ln k} = \frac{e^{\ln k} - e^0}{\ln k} = \frac{k-1}{\ln k}$ . 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, $k=e$ .
What is the average value of $f(x) = \ln(x)$ on the interval $[1, e]$ ?
Answer: $\frac{1}{e-1}$ The average value is $\frac{1}{e-1} \int_1^e \ln(x) \,dx$ . The antiderivative of $\ln(x)$ is $x \ln(x) - x$ . Evaluating the definite integral: $[x \ln(x) - x]_1^e = (e \ln e - e) - (1 \ln 1 - 1) = (e - e) - (0 - 1) = 1$ . The average value is $\frac{1}{e-1} \cdot 1 = \frac{1}{e-1}$ . Distractor A is the value of the integral. Distractor B incorrectly multiplies the integral by the interval length. Distractor C results from an error in finding the antiderivative.
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 $t=6$ )?
Answer: $68 + \frac{8}{\pi}$ 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 $t=0$ to $t=6$ : $\text{Average} = \frac{1}{6-0}\int_0^6 \left[68 + 4\sin(\frac{\pi t}{12})\right]dt$ $= \frac{1}{6}\left[\int_0^6 68\,dt + \int_0^6 4\sin(\frac{\pi t}{12})\,dt\right]$ 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.
The average value of the derivative of a differentiable function, $f'(x)$ , on an interval $[a, b]$ represents which quantity?
Answer: The average rate of change of the function $f(x)$ on $[a, b]$ . The average value of $f'(x)$ on $[a, b]$ 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 $[a, b]$ .
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 $k$ ?
Answer: $4$ 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 $[0, 3]$ . 1. Set up the equation based on the formula: $13 = \frac{1}{3-0} \int_0^3 (kx^2 + 1) \, dx$ 2. Calculate the definite integral: $\int_0^3 (kx^2 + 1) \, dx = \left[ \frac{k}{3}x^3 + x \right]_0^3 = \left( \frac{k}{3}(3)^3 + 3 \right) - (0) = 9k + 3$ 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 $13 = 3k+1$ .
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 $f_{avg} \cdot g_{avg}$ ?
Answer: $8/\pi^2$ 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 $\sec^2(x)$ on $[0, \pi/4]$ . 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 $\cos(x)$ on $[0, \pi/2]$ . 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.
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 $k$ ?
Answer: $1/2$ 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 $f(x) = 3x^2 - 2x$ on $[0, k]$ 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.
What is the average value of the function $f(x) = x\sqrt{x^2+9}$ on the interval $[0, 4]$ ?
Answer: $49/6$ 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 $f(x) = x\sqrt{x^2+9}$ on $[0,4]$ , 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.
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?
Answer: $135^{\circ}C$ The average temperature is the average value of the function $T(x)$ on $[0, 3]$ . $T_{avg} = \frac{1}{3-0} \int_0^3 (270x - 90x^2) \,dx = \frac{1}{3} [135x^2 - 30x^3]_0^3 = \frac{1}{3} [135(3^2) - 30(3^3)] = \frac{1}{3}[1215 - 810] = \frac{405}{3} = 135$ . 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.
Let $A(h)$ represent the average value of the function $f(x) = e^x$ on the interval $[2, 2+h]$ . Find $\lim_{h \to 0} A(h)$ .
Answer: $e^2$ The average value is $A(h) = \frac{1}{(2+h)-2} \int_2^{2+h} e^x \,dx = \frac{1}{h} [e^x]_2^{2+h} = \frac{e^{2+h} - e^2}{h}$ . The limit $\lim_{h \to 0} \frac{e^{2+h} - e^2}{h}$ is the definition of the derivative of the function $g(x) = e^x$ at $x=2$ . Since $g'(x) = e^x$ , the limit is $g'(2) = e^2$ . Conceptually, as the interval shrinks, the average value of a continuous function approaches the function's value at that point.
Let $g(x) = \int_1^x t^2 \,dt$ . What is the average value of the function $g(x)$ on the interval $[1, 4]$ ?
Answer: $\frac{27}{4}$ This problem combines two fundamental calculus concepts: the Fundamental Theorem of Calculus and average value of functions. When you see a function defined as an integral with a variable upper limit, you'll need to first find an explicit formula for that function. Start by evaluating the integral $g(x) = \int_1^x t^2 \,dt$ . Using the antiderivative of $t^2$ , which is $\frac{t^3}{3}$ , you get: $g(x) = \left[\frac{t^3}{3}\right]_1^x = \frac{x^3}{3} - \frac{1^3}{3} = \frac{x^3 - 1}{3}$ Now find the average value using the formula: $\text{Average} = \frac{1}{b-a}\int_a^b g(x)\,dx$ For the interval $[1,4]$ : $\text{Average} = \frac{1}{4-1}\int_1^4 \frac{x^3-1}{3}\,dx = \frac{1}{9}\int_1^4 (x^3-1)\,dx$ $= \frac{1}{9}\left[\frac{x^4}{4} - x\right]_1^4 = \frac{1}{9}\left[\left(\frac{256}{4} - 4\right) - \left(\frac{1}{4} - 1\right)\right]$ $= \frac{1}{9}(64 - 4 - 0.25 + 1) = \frac{1}{9} \cdot \frac{243}{4} = \frac{27}{4}$ Choice A is correct. Choice B ( $\frac{85}{12}$ ) likely results from computational errors in the integration. Choice C (7) might come from forgetting the coefficient $\frac{1}{3}$ in $g(x)$ . Choice D ( $\frac{243}{4}$ ) is what you'd get if you forgot to divide by the interval length. Remember: when finding average values of functions defined by integrals, always evaluate the integral first to get an explicit function, then apply the average value formula carefully.
Let $f$ be a continuous function and let $F$ be an antiderivative of $f$ . The average value of $f$ on the interval $[a, b]$ is equivalent to which of the following quantities?
Answer: The average rate of change of $F(x)$ on $[a, b]$ . When you see a question connecting average values with antiderivatives, think about how these concepts relate through the Fundamental Theorem of Calculus and the Mean Value Theorem. The average value of a function $f$ on $[a,b]$ is defined as $\frac{1}{b-a}\int_a^b f(x)\,dx$ . Since $F$ is an antiderivative of $f$ , the Fundamental Theorem tells us that $\int_a^b f(x)\,dx = F(b) - F(a)$ . Therefore, the average value becomes $\frac{F(b) - F(a)}{b - a}$ . This expression is precisely the average rate of change of $F(x)$ on $[a,b]$ - the change in $F$ divided by the change in $x$ . So answer D is correct. Let's see why the other choices fail. Choice A suggests averaging the endpoint values of $F$ , but this ignores the behavior of $F$ throughout the interval and has no connection to integration. Choice B describes a specific point $c$ where $f(c)$ equals the total change in $F$ , but this represents one particular value of $f$ , not its average over the entire interval. Choice C refers to the average rate of change of $f$ itself, which would be $\frac{f(b) - f(a)}{b - a}$ - completely different from the average value of $f$ . Remember this key connection: the average value of a function equals the average rate of change of its antiderivative. This relationship appears frequently in calculus problems linking integration and differentiation.
The function $F(x) = x^3 - \cos(x)$ is an antiderivative of a function $f(x)$ . What is the average value of $f(x)$ on the interval $[0, \pi]$ ?
Answer: $\pi^2 + \frac{2}{\pi}$ This question combines two fundamental calculus concepts: the relationship between functions and their antiderivatives, and the formula for average value of a function. Since $F(x) = x^3 - \cos(x)$ is an antiderivative of $f(x)$ , you need to find $f(x)$ by taking the derivative: $f(x) = F'(x) = 3x^2 + \sin(x)$ . Remember that the derivative of $\cos(x)$ is $-\sin(x)$ , so the derivative of $-\cos(x)$ is $\sin(x)$ . The average value of a function on interval $[a,b]$ is $\frac{1}{b-a}\int_a^b f(x)\,dx$ . Here, you need $\frac{1}{\pi-0}\int_0^\pi (3x^2 + \sin(x))\,dx = \frac{1}{\pi}\int_0^\pi (3x^2 + \sin(x))\,dx$ . Since $F(x)$ is an antiderivative of $f(x)$ , the Fundamental Theorem of Calculus gives us: $\int_0^\pi f(x)\,dx = F(\pi) - F(0)$ . Calculate: $F(\pi) = \pi^3 - \cos(\pi) = \pi^3 - (-1) = \pi^3 + 1$ and $F(0) = 0^3 - \cos(0) = 0 - 1 = -1$ . So $\int_0^\pi f(x)\,dx = (\pi^3 + 1) - (-1) = \pi^3 + 2$ . The average value is $\frac{\pi^3 + 2}{\pi} = \pi^2 + \frac{2}{\pi}$ , which is answer C. Answer A adds $\frac{1}{\pi}$ instead of $\frac{2}{\pi}$ , likely from miscalculating the definite integral. Answer B forgets the $\sin(x)$ term entirely. Answer D gives the value of the definite integral itself, not the average value. Key strategy: When given an antiderivative, always differentiate first to find the original function, then use the Fundamental Theorem to evaluate definite integrals efficiently.
Find the smallest positive value of $c$ in the interval $[0, 2\pi]$ such that $f(c)$ equals the average value of $f(x) = \sin(x)$ on $[0, 2\pi]$ .
Answer: $\pi$ First, find the average value of $\sin(x)$ on $[0, 2\pi]$ . $f_{avg} = \frac{1}{2\pi} \int_0^{2\pi} \sin(x) \,dx = \frac{1}{2\pi} [-\cos(x)]_0^{2\pi} = \frac{1}{2\pi} (-\cos(2\pi) - (-\cos(0))) = \frac{1}{2\pi}(-1 - (-1)) = 0$ . Next, we must find values of $c$ in $[0, 2\pi]$ for which $f(c) = 0$ , i.e., $\sin(c) = 0$ . The solutions are $c = 0, \pi, 2\pi$ . The smallest positive value among these is $\pi$ .
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?
Answer: $36$ 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 $\frac{2}{3}x^{3/2}$ .
Let $c$ be a value in the interval $[0, 4]$ guaranteed by the Mean Value Theorem for Integrals for the function $f(x) = 3x^2 - 6x$ . What is the value of $c$ ?
Answer: $1 + \frac{\sqrt{21}}{3}$ First, find the average value of the function on the interval $[0, 4]$ : $f_{avg} = \frac{1}{4-0} \int_0^4 (3x^2 - 6x) \,dx = \frac{1}{4} [x^3 - 3x^2]_0^4 = \frac{1}{4}(64 - 48) = \frac{16}{4} = 4$ . According to the Mean Value Theorem for Integrals, there exists a $c$ in $[0, 4]$ such that $f(c) = f_{avg}$ . We must solve $3c^2 - 6c = 4$ , which is $3c^2 - 6c - 4 = 0$ . Using the quadratic formula, $c = \frac{6 \pm \sqrt{36 - 4(3)(-4)}}{6} = \frac{6 \pm \sqrt{84}}{6} = 1 \pm \frac{2\sqrt{21}}{6} = 1 \pm \frac{\sqrt{21}}{3}$ . The two possible values are $c \approx 1 + 1.53 = 2.53$ and $c \approx 1 - 1.53 = -0.53$ . Only $1 + \frac{\sqrt{21}}{3}$ is in the interval $[0, 4]$ .
What is the average value of the function $f(x) = \cos^2(x)$ on the interval $[0, \pi/2]$ ?
Answer: $\frac{1}{2}$ To find the average value, we compute $\frac{1}{\pi/2 - 0} \int_0^{\pi/2} \cos^2(x) \,dx$ . We use the half-angle identity $\cos^2(x) = \frac{1 + \cos(2x)}{2}$ . The expression becomes $\frac{2}{\pi} \int_0^{\pi/2} \frac{1 + \cos(2x)}{2} \,dx = \frac{1}{\pi} [x + \frac{1}{2}\sin(2x)]_0^{\pi/2} = \frac{1}{\pi} [(\frac{\pi}{2} + \frac{1}{2}\sin(\pi)) - (0 + \frac{1}{2}\sin(0))] = \frac{1}{\pi}(\frac{\pi}{2}) = \frac{1}{2}$ . Distractor A is the value of the integral. Distractor B is the average value of $\cos(x)$ on the interval.
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 $c$ ?
Answer: $e-1$ The Mean Value Theorem for Integrals states that if $f$ is continuous on $[a, b]$ , there exists a $c$ in $[a, b]$ such that $f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx$ . 1. First, calculate the average value of $f(x) = 1/x$ on $[1, e]$ : $f_{avg} = \frac{1}{e-1} \int_1^e \frac{1}{x} \, dx$ 2. The integral of $1/x$ is $\ln|x|$ : $\int_1^e \frac{1}{x} \, dx = [\ln|x|]_1^e = \ln(e) - \ln(1) = 1 - 0 = 1$ 3. Substitute this back into the average value formula: $f_{avg} = \frac{1}{e-1} \cdot 1 = \frac{1}{e-1}$ 4. Now, find the value of $c$ such that $f(c) = f_{avg}$ : $f(c) = \frac{1}{c} = \frac{1}{e-1}$ 5. 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.

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