📝 Mean value theorem for integrals (29 MCQs)

📖 Explanation: The theorem states that there is at least one point $x^*$ in $[a,b]$ such that the integral of $f$ from $a$ to $b$ equals $f(x^*)(b-a)$. This means the area under the curve is equal to the area of a rectangle whose base is the interval $[a,b]$ and whose height is the function's value at some point $x^*$. This height represents the average value of the function over the interval. The rectangle's area is precisely the area under the curve, providing a powerful bridge between integration and geometry.

📖 Explanation: The Mean-Value Theorem for Integrals is fundamentally linked to the concept of the average value of a continuous function. The theorem asserts that the number $\frac{1}{b-a}\int_a^b f(x) dx$, which is defined as the average value of $f$ over $[a,b]$, is actually a value that the function $f$ takes on at some point $x^*$ within the interval. This is a deep result, as it confirms that the 'average' of infinitely many function values is not just an abstract number but a concrete value achieved by the function itself.

📖 Explanation: The average value of a function $f$ on an interval $[a,b]$ is defined as $f_{avg} = \frac{1}{b-a}\int_a^b f(x) dx$. In this problem, $a=2$, $b=5$, and $f_{avg}=7$. Substituting these values into the definition, we get $7 = \frac{1}{5-2}\int_2^5 f(x) dx = \frac{1}{3}\int_2^5 f(x) dx$. Solving for the integral, we get $\int_2^5 f(x) dx = 3 \times 7 = 21$.

📖 Explanation: The Mean-Value Theorem for Integrals specifically requires the function $f$ to be continuous on the closed interval $[a,b]$. While differentiability is a condition for the Mean Value Theorem for Derivatives, it is not required here. The theorem relies on the Extreme Value Theorem and the Intermediate Value Theorem, both of which require continuity. The theorem applies to a broad class of continuous functions, not just polynomials or monotonic functions.

📖 Explanation: First, find the average value of $f(x)=x^2$ on $[0,2]$: $f_{avg} = \frac{1}{2-0}\int_0^2 x^2 dx = \frac{1}{2} [\frac{x^3}{3}]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$. The theorem guarantees a point $x^*$ where $f(x^*) = f_{avg}$, so $(x^*)^2 = \frac{4}{3}$. Solving for $x^*$ in $[0,2]$ gives $x^* = \frac{2}{\sqrt{3}}$. The student's answer of $2/3$ is incorrect. The error likely stems from incorrectly evaluating the integral or setting up the equation $f(x^*) = f_{avg}$.

📖 Explanation: The average value of $f$ on $[1,4]$ is $f_{avg} = \frac{1}{4-1}\int_1^4 3x^2 dx = \frac{1}{3} [x^3]_1^4 = \frac{1}{3}(64-1) = 21$. According to the theorem, there exists a point $x^*$ such that $f(x^*) = f_{avg}$, so $3(x^*)^2 = 21$, which gives $(x^*)^2 = 7$. Therefore, $x^* = \pm \sqrt{7}$. Since $x^*$ must lie in the interval $[1,4]$, the valid point is $x^* = \sqrt{7}$.

📖 Explanation: For a constant function $f(x) = 5$, the function's value is 5 at every point in the interval. The theorem states there exists at least one point $x^*$ such that $f(x^*) = f_{avg}$. The average value of a constant function is that constant value, $f_{avg}=5$. Therefore, for any point $x^*$ in $[2,7]$, $f(x^*) = 5 = f_{avg}$. The theorem does not guarantee uniqueness; in fact, every point in the interval satisfies the condition.

📖 Explanation: The Mean-Value Theorem for Integrals states that there exists a time $t^*$ such that $\int_0^4 v(t) dt = v(t^*)(4-0)$. The integral $\int_0^4 v(t) dt$ is the displacement of the particle. Thus, the displacement is equal to the product of the instantaneous velocity at time $t^*$ and the total time elapsed. In other words, the particle's displacement is the same as if it had been moving at the constant velocity $v(t^*)$ for the entire 4 seconds. This theorem guarantees the existence of this 'average' velocity as an actual velocity experienced by the particle.

📖 Explanation: The average value is $f_{avg} = \frac{1}{5-0}\int_0^5 f(x) dx = \frac{1}{5} \times 10 = 2$. This means the rectangle with base $[0,5]$ and height 2 has an area of 10, which is the same as the area under the curve. This is the geometric interpretation, so option B is correct. Also, the Mean-Value Theorem for Integrals guarantees that there is a point $x^*$ where $f(x^*) = f_{avg}$. Since $f(x^*)$ is a value of the function, it must lie between the minimum and maximum values of $f$ on the interval. Thus, option A is also correct.

📖 Explanation: The Mean-Value Theorem for Integrals requires the function $f$ to be continuous on the closed interval $[a,b]$. The function $f(x)=1/x$ is not defined at $x=0$, which lies within the interval $[-1,1]$. Therefore, $f$ is not continuous on the interval, and the theorem cannot be applied. This is a common error; students may mistakenly apply the theorem to functions with discontinuities or jump discontinuities, leading to incorrect conclusions about the existence of an $x^*$. The theorem does not require differentiability, but continuity is an absolute necessity.

📖 Explanation: The average value of $f(x)=mx+c$ over $[a,b]$ is $f_{avg} = \frac{1}{b-a} \int_a^b (mx+c) dx = \frac{1}{b-a} [\frac{m}{2}x^2 + cx]_a^b = \frac{m(a+b)}{2} + c$. For a linear function, the value at the midpoint is $f(\frac{a+b}{2}) = m(\frac{a+b}{2}) + c = \frac{m(a+b)}{2} + c$, which equals $f_{avg}$. Thus, the point $x^*$ is always the midpoint of the interval, $x^* = \frac{a+b}{2}$. This is a special property of linear functions and aligns with the geometric intuition that the area under a line is the area of a rectangle with height equal to the line's value at its midpoint.

📖 Explanation: First, find the value of the original integral: $\int_{-2}^3 f(x) dx = f_{avg} \times (b-a) = 4 \times (3-(-2)) = 4 \times 5 = 20$. Then, use the linearity property of integrals: $\int_{-2}^3 (f(x) + 1) dx = \int_{-2}^3 f(x) dx + \int_{-2}^3 1 dx = 20 + [x]_{-2}^3 = 20 + (3-(-2)) = 20 + 5 = 25$. This question tests the understanding of how to combine the definition of average value with the basic properties of definite integrals.

📖 Explanation: The average value of a function over a larger interval does not provide specific information about the integral over a subinterval. Knowing that $\int_0^6 f(x) dx = 10 \times 6 = 60$ tells us the total area under the curve over the entire interval. However, without knowing the shape of the function or its behavior on the subinterval $[0,3]$, we cannot determine the exact area under the curve on that specific part. It could be any value between 0 and 60, depending on how the function's values are distributed across the interval. This is a common misconception where students assume a uniform distribution of the function's values.

📖 Explanation: This question tests the distinction between the Mean Value Theorem for Derivatives (which requires differentiability) and the Mean-Value Theorem for Integrals (which requires only continuity). The function $f(x)=|x|$ is continuous on $[-1,2]$; it does not have any jumps, breaks, or asymptotes. Although it is not differentiable at $x=0$, this is irrelevant for the integral version of the theorem. The theorem will guarantee the existence of some point $x^*$ where the function's value equals its average value over the interval. The student's error is in applying the wrong set of conditions.

📖 Explanation: If $\int_a^b f(x) dx = 0$, then the average value of $f$ on $[a,b]$ is $f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx = 0$. The Mean-Value Theorem for Integrals guarantees that there is a point $x^*$ where $f(x^*) = f_{avg}$. Since $f_{avg}=0$, we have $f(x^*) = 0$ for at least one point $x^*$ in $[a,b]$. This is a powerful consequence; the integral of a continuous function being zero implies that the function itself must be zero somewhere on the interval. It does not mean the function is identically zero; for example, $f(x)=2x-1$ on $[0,1]$ has an integral of 0 but is not zero everywhere.

📖 Explanation: The average profit rate over the interval $[0,4]$ is $\frac{1}{4-0}\int_0^4 (3t^2 - 12t + 9) dt = \frac{1}{4} [t^3 - 6t^2 + 9t]_0^4 = \frac{1}{4} (64 - 96 + 36) = \frac{1}{4} (4) = 1$. Wait, calculation: $[t^3 - 6t^2 + 9t]_0^4 = (64 - 96 + 36) - 0 = 4$. So average = $\frac{1}{4} \times 4 = 1$. However, we need to check the options. Let's re-evaluate: $\int_0^4 (3t^2 - 12t + 9) dt = [t^3 - 6t^2 + 9t]_0^4 = (64 - 96 + 36) = 4$. Average = $4/4 = 1$. The correct answer is $1$. But there is a mistake. Let's re-read the problem. P'(t) = 3t^2 - 12t + 9. The average is $\frac{1}{4} \int_0^4 (3t^2 - 12t + 9) dt = \frac{1}{4} [t^3 - 6t^2 + 9t]_0^4 = \frac{1}{4}(64 - 96 + 36) = \frac{1}{4}(4)=1$. So option B is 1. But the option A is 3. Let's correct: Option B is 1. However, the calculation is correct. Let's change option A to 3, B to 1, C to 9, D to 12. So the answer should be B.

📖 Explanation: For an even function $f$ on $[-a,a]$, the integral is $\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx$. The average value is $\frac{1}{2a} \int_{-a}^a f(x) dx = \frac{1}{a}\int_0^a f(x) dx$. The theorem guarantees a point $x^*$ such that $f(x^*) = \frac{1}{a}\int_0^a f(x) dx$. However, the theorem does not specify where this point is; it could be positive, negative, or zero. The symmetry of the function doesn't force the point to be at the origin or in pairs. The only certain conclusion is about the value of $f(x^*)$, which is the average value of the function over the interval.

📖 Explanation: The Mean-Value Theorem for Integrals is a distinct theorem that provides a powerful geometric interpretation of the definite integral. It states that for a continuous function $f$ on $[a,b]$, there exists a number $x^*$ in $[a,b]$ such that $\int_a^b f(x) dx = f(x^*)(b-a)$. The right-hand side is the area of a rectangle with base $b-a$ and height $f(x^*)$. The theorem proves that there is always a rectangle with the same base as the interval whose area is exactly equal to the area under the curve. This is a direct statement about the existence of such a rectangle.

📖 Explanation: The average value of $f(x)=\sqrt{x}$ on $[0,4]$ is $f_{avg} = \frac{1}{4-0}\int_0^4 \sqrt{x} dx = \frac{1}{4} [\frac{2}{3} x^{3/2}]_0^4 = \frac{1}{4} \cdot \frac{2}{3} \cdot 8 = \frac{4}{3}$. According to the theorem, we need to find $x^*$ such that $f(x^*) = f_{avg}$, so $\sqrt{x^*} = \frac{4}{3}$. Squaring both sides gives $x^* = \frac{16}{9}$. It is important to verify that $x^*$ lies in the interval $[0,4]$. Since $\frac{16}{9} \approx 1.78$, it is within the interval, so it is a valid solution.

📖 Explanation: The average value formula is $T_{avg} = \frac{1}{12-0}\int_0^{12} T(t) dt = \frac{1}{12}\int_0^{12} T(t) dt$. We are given that $T_{avg} = 20$. Therefore, $\frac{1}{12}\int_0^{12} T(t) dt = 20$. Solving for the integral, we get $\int_0^{12} T(t) dt = 20 \times 12 = 240$. The units are $^\circ C \times \text{hours}$, which is the correct unit for the integral of temperature over time. This integral represents the total 'temperature-hours' or the accumulated thermal effect over the 12-hour period.

📖 Explanation: If $f(x) \ge g(x)$ on $[1,5]$, then $\int_1^5 f(x) dx \ge \int_1^5 g(x) dx$. Dividing both sides by the positive length of the interval $(5-1)=4$ maintains the inequality: $\frac{1}{4}\int_1^5 f(x) dx \ge \frac{1}{4}\int_1^5 g(x) dx$, which means $f_{avg} \ge g_{avg}$. Option A states this. Additionally, if the integrals are equal, then the average values are equal. The inequality $f(x) \ge g(x)$ does not necessarily mean $f_{avg} > g_{avg}$ because the functions could be equal at all points or have equal integrals despite being different. Option C is possible only if the integrals are equal. Therefore, both A and C are possible outcomes, making D the most complete answer.

📖 Explanation: By the Mean-Value Theorem for Integrals, since speed is a continuous function of time, there exists at least one instant $t^*$ in the time interval during which the instantaneous speed equals the average speed of 50 mph. This is a direct Easy of the theorem. The theorem does not imply that the speed was constant at 50 mph; only that there was at least one moment when the speedometer read exactly 50 mph. The car could have varied its speed above and below 50 mph, but the average speed over the entire trip being 50 mph guarantees that at least one point in time the speed was exactly 50 mph.

📖 Explanation: The average value of $\sin x$ on $[0,\pi]$ is $f_{avg} = \frac{1}{\pi-0}\int_0^\pi \sin x dx = \frac{1}{\pi} [-\cos x]_0^\pi = \frac{1}{\pi} (-\cos \pi + \cos 0) = \frac{1}{\pi} (1+1) = \frac{2}{\pi}$. The theorem guarantees a point $x^*$ such that $f(x^*) = f_{avg}$, so $\sin(x^*) = \frac{2}{\pi}$. The solution for $x^*$ in $[0,\pi]$ is $x^* = \sin^{-1}(\frac{2}{\pi})$. Note that $\sin^{-1}(\frac{2}{\pi})$ is approximately 0.69, which is in the interval. The average value is not $\frac{1}{2}$, and $x^*$ is not simply $\pi/2$ because $\sin(\pi/2)=1$.

📖 Explanation: The average value of $x^3$ on $[-2,2]$ is indeed 0 because it's an odd function integrated over a symmetric interval. The student correctly finds that $f(0)=0$, so $x^*=0$ is a point where the function equals its average value. However, the Mean-Value Theorem for Integrals only guarantees the existence of at least one point. For $f(x)=x^3$, there is only one point where $f(x)=0$ on $[-2,2]$, which is $x=0$. So the student's conclusion is correct in this case, but the reasoning 'because the average is 0, $x^*=0$' is not generally valid. For other functions, there could be multiple points. The student's reasoning is valid for this specific case because the function is one-to-one. The error is assuming that $x^*$ is the only point. However, in this case, it is the only point. The question is tricky. Let's say the student says $x^*=0$ is the point. That is correct. But the reasoning is not generally valid. The answer should be C. But if the student says $x^*=0$ is the only point, that is true for $x^3$. The question asks if the reasoning is valid. The reasoning 'I found the average is 0, so $x^*=0$' is not a valid logical step for all functions. For this function, it happens to be correct. The most accurate criticism is that the student has not proven uniqueness and their method of simply setting $x$ to 0 is not a general procedure. The theorem doesn't require uniqueness. The correct answer is C.

📖 Explanation: The Mean-Value Theorem for Integrals applies to functions that are continuous on a closed, finite interval $[a,b]$. Among the options, only option C states a continuous function on $[-2,2]$, which is a closed, finite interval. Option A has a discontinuity at 0, so the theorem does not apply. Option B is an infinite interval, and the theorem is not defined for such intervals in its basic form. Option D has a jump discontinuity, which violates the continuity requirement. Therefore, only the function in option C is guaranteed to have such a point $x^*$.

📖 Explanation: The average value of a function is defined as $P_{avg} = \frac{1}{b-a}\int_a^b P(t) dt$. Here, $a=0$, $b=24$ (hours), and $P_{avg} = 35$ (micrograms per cubic meter). The total exposure is the integral $\int_0^{24} P(t) dt = P_{avg} \times (24-0) = 35 \times 24 = 840$. The units are (micrograms per cubic meter) multiplied by hours, which gives 'microgram-hours per cubic meter.' This is the correct unit for the integral, representing the cumulative exposure over time. Option B is the only one with the correct numerical value and units.

📖 Explanation: The function is $f(x) = 2x$ on $[0,4]$. The average value is $\frac{1}{4}\int_0^4 2x dx = \frac{1}{4} [x^2]_0^4 = \frac{1}{4}(16) = 4$. We need $f(x^*) = 4$, so $2x^* = 4$, which gives $x^* = 2$. Geometrically, the area under the line $y=2x$ from 0 to 4 is a triangle with area $\frac{1}{2} \times 4 \times 8 = 16$. A rectangle with base 4 and height 4 has the same area. The height of this rectangle is $f(2) = 4$, so the rectangle's top is the point on the line at $x=2$. This makes sense because the midpoint of a line's interval always gives the average value.

📖 Explanation: The average value is $f_{avg} = \frac{1}{10-0}\int_0^{10} f(x) dx = \frac{50}{10} = 5$. The Mean-Value Theorem for Integrals guarantees that there exists an $x^*$ in $[0,10]$ such that $f(x^*) = f_{avg} = 5$. This is the fundamental result of the theorem. The other options are incorrect because they do not represent the average value of the function. This question tests the direct Easy of the theorem's conclusion: the function attains its average value at some point. The integral value of 50 and interval length 10 directly give the average value of 5.

📖 Explanation: The average value is $f_{avg} = \frac{1}{1-0}\int_0^1 e^x dx = e - 1$. The theorem states there is a point $x^*$ such that $f(x^*) = f_{avg}$, so $e^{x^*} = e - 1$. This is correct. Now, find $x^*$: $x^* = \ln(e-1)$. Since $e \approx 2.718$, $e-1 \approx 1.718$, and $\ln(1.718) \approx 0.541$. This value is in the interval $[0,1]$, so the point exists. The student's statement is correct. However, the student might not have realized that $x^*$ is specifically $\ln(e-1)$. The theorem guarantees existence, and the point is in the interval. So the statement is correct. But let's check option D: It says 'No, the theorem guarantees $e^{x^*} = e - 1$, which is correct, but $x^*$ is not in $[0,1]$.' This is false because $\ln(e-1)$ is in $[0,1]$. The correct answer is C, but C says 'Yes, but the point $x^*$ is $\ln(e-1)$.' That is the most precise. Let's correct the options. The correct answer is C.