Answer: The peer is correct; $\int_5^2 (4x)\,dx = -\int_2^5 (4x)\,dx = -42$.

Verify $\int_2^5 4x\,dx$: $\int_2^5 4x\,dx = [2x^2]_2^5 = 2(25) - 2(4) = 50 - 8 = 42$. Apply the reversal property: $\int_5^2 4x\,dx = -\int_2^5 4x\,dx = -42$. The peer is correct. Why distractors fail: Option A ignores the reversal-of-limits property. Option C has no basis. Option D is wrong because the definite integral is well-defined even when the lower limit exceeds the upper limit.

Answer: $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x$

Formal definition of the definite integral: The definite integral is defined as the limit of Riemann sums as the number of subintervals $n$ approaches infinity: $\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x$. Why the correct answer works: Option B matches the formal definition exactly — the limit of the Riemann sum as $n \to \infty$. Why distractors fail: Option A is a Riemann sum approximation without the limit. Option C takes $n \to 0$, which is meaningless in this context. Option D applies a derivative to the sum, which is not part of the definition.

Answer: Narrower subintervals allow the rectangles to follow the shape of the curve more closely, reducing the gap between the rectangles and the actual curve.

Core intuition: As $n$ increases, $\Delta x = \frac{b-a}{n}$ decreases, making each rectangle narrower. Narrower rectangles conform more tightly to the curve, reducing the approximation error. Why the correct answer works: Option B captures the essential geometric reason: narrower rectangles better match the curvature, decreasing the total error between the Riemann sum and the true area. Why distractors fail: Option A incorrectly claims rectangles get taller; their height depends on the function, not on $n$. Option C is false — finite Riemann sums are almost always approximations. Option D confuses function values with accuracy; more evaluation points don't make the function values larger.

Answer: $\sum_{i=1}^{n} \sqrt{1 + \frac{4i}{n}} \cdot \frac{4}{n}$

Set up $\Delta x$ and right endpoints: $\Delta x = \frac{5-1}{n} = \frac{4}{n}$. The right endpoint of the $i$-th subinterval is $x_i = 1 + i\,\Delta x = 1 + \frac{4i}{n}$. Write the sigma notation: The right Riemann sum is $R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} \sqrt{1 + \frac{4i}{n}} \cdot \frac{4}{n}$. Why distractors fail: Option B uses $i = 0$ to $n-1$, which gives left endpoints. Option C incorrectly starts the interval at 0 instead of 1 and uses $\Delta x = 5/n$. Option D uses the correct starting point but wrong interval length ($5$ instead of $4$).

Answer: The trapezoidal approximation

Effect of concavity on approximation methods: For a concave up function ($f'' > 0$), the curve lies below the secant line connecting endpoints of each subinterval. The trapezoidal rule uses these secant lines, so it captures more area than is actually under the curve. Midpoint behavior for concave up functions: Conversely, for concave up functions, the tangent line at the midpoint lies below the curve, so the midpoint approximation underestimates. Why distractors fail: Option A is wrong because the midpoint rule underestimates for concave up functions. Option C is wrong because only the trapezoidal rule overestimates. Option D is wrong because concavity does determine the direction of the error for both methods.

Answer: $\Delta x = \frac{b - a}{n}$

Partitioning an interval: When dividing $[a, b]$ into $n$ equal subintervals, the total length of the interval is $b - a$. Each subinterval has width $\Delta x = \frac{b - a}{n}$. Why the correct answer works: $\Delta x = \frac{b - a}{n}$ correctly divides the total interval length by the number of subintervals. Why distractors fail: Option A uses $a + b$ instead of the interval length $b - a$. Option C divides by $n + 1$, which confuses the number of subintervals with the number of partition points. Option D inverts the fraction.

Answer: $52$ m

Set up the trapezoidal rule: $\Delta t = 2$ seconds, with partition points at $t = 0, 2, 4, 6$. The trapezoidal rule gives: $T = \frac{\Delta t}{2}[v(0) + 2v(2) + 2v(4) + v(6)]$. Substitute values: $T = \frac{2}{2}[0 + 2(5) + 2(12) + 18] = 1 \cdot [0 + 10 + 24 + 18] = 52$ m. Why distractors fail: Option A ($35$) could come from a left Riemann sum: $2(0 + 5 + 12) = 34$ (close but incorrect). Option C ($70$) might result from doubling the sum incorrectly. Option D ($58$) could come from a right Riemann sum: $2(5 + 12 + 18) = 70$ — also incorrect.

Answer: $0$

Interval of zero width: When the upper and lower limits of integration are the same, the interval has zero width, so there is no area to accumulate. Why the correct answer works: By the definition of the definite integral, $\int_a^a f(x)\,dx = 0$ because $\Delta x = \frac{a - a}{n} = 0$. Why distractors fail: Option A confuses the integral with a function evaluation. Option B has no basis in the definition. Option D is incorrect because the integral is perfectly well-defined — it simply equals zero.

Answer: $\frac{x_{i-1} + x_i}{2}$

Midpoint selection: The midpoint Riemann sum evaluates the function at the midpoint of each subinterval, which is the arithmetic mean of the two endpoints: $x_i^* = \frac{x_{i-1} + x_i}{2}$. Why the correct answer works: The midpoint of $[x_{i-1}, x_i]$ is $\frac{x_{i-1} + x_i}{2}$, which is the standard definition of the midpoint rule. Why distractors fail: Option A is the left endpoint. Option B is the right endpoint. Option D uses the product instead of the sum of the endpoints, which does not give the midpoint.

Answer: $12$

Determine $\Delta x$ and partition points: $\Delta x = \frac{3-0}{3} = 1$. The partition points are $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3$. Evaluate the function: $f(0) = 1$, $f(1) = 3$, $f(2) = 5$, $f(3) = 7$. Apply the trapezoidal rule: $T_3 = \frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3)] = \frac{1}{2}[1 + 6 + 10 + 7] = \frac{24}{2} = 12$. Why distractors fail: Option A ($9$) omits a term or misapplies the formula. Option C ($15$) may come from using the right Riemann sum without halving. Option D ($10$) could result from a computational error in the function evaluations.

Answer: No, both methods are exact for constant functions, and for certain non-constant functions the left sum can match or beat the trapezoidal rule due to favorable error cancellation.

Analyze the claim: In general, the trapezoidal rule has a smaller error bound ($O(1/n^2)$) than the left Riemann sum ($O(1/n)$). However, "always more accurate" is too strong a claim. Find counterexamples: For constant functions $f(x) = c$, both methods give the exact answer, so the trapezoidal rule is not "more accurate." Additionally, for specific non-constant functions and particular values of $n$, the left Riemann sum's errors can cancel across subintervals, producing an estimate that is as accurate as or even closer to the true value than the trapezoidal estimate. Why distractors fail: Options A and B assert that the trapezoidal rule is always better, ignoring cases where both are exact or where error cancellation favors the left sum. Option D goes too far in the other direction — the trapezoidal rule generally performs at least as well as the left Riemann sum, but not always strictly better.

Answer: The student forgot to multiply the sum by $\Delta x$.

Identify the missing factor: The right Riemann sum is $R_n = \Delta x \sum_{i=1}^{n} f(x_i)$. The student correctly identified the right endpoints but did not multiply the sum by $\Delta x = \pi/4$. Check the endpoints: With $n = 4$ and $\Delta x = \pi/4$, the right endpoints are $x_1 = \pi/4$, $x_2 = \pi/2$, $x_3 = 3\pi/4$, $x_4 = \pi$. These are correct. Why distractors fail: Option B is wrong because the endpoints are correct. Option C would give 5 subintervals, not 4. Option D is wrong because the multiplication by $\Delta x$ is essential and was omitted.

Answer: $9$

Identify the integral: The sum has $\Delta x = \frac{3}{n}$ and $x_i = \frac{3i}{n}$, so $a = 0$, $b = 3$, and $f(x) = x^2$. The limit equals $\int_0^3 x^2\,dx$. Evaluate the integral: $\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9$. Why distractors fail: Option A ($3$) might come from evaluating $\frac{x^3}{3}$ at $x=3$ without cubing first. Option C ($27$) forgets to divide by 3 in the antiderivative. Option D ($18$) could result from using $\int_0^3 2x^2\,dx$ by mistake.

Answer: $30$

Determine subinterval width: $\Delta x = \frac{4-0}{4} = 1$. The right endpoints are $x = 1, 2, 3, 4$. Evaluate the function at right endpoints: $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, $f(4) = 16$. Compute the sum: $R_4 = \Delta x\,[f(1) + f(2) + f(3) + f(4)] = 1(1 + 4 + 9 + 16) = 30$. Why distractors fail: Option A ($14$) is the left Riemann sum $L_4 = 1(0 + 1 + 4 + 9)$. Option C ($20$) could result from an averaging error. Option D ($64/3$) is the exact value of the integral, not the Riemann sum approximation.

Answer: Yes, because for a linear function the midpoint value equals the average value on each subinterval, making each rectangle's area exactly equal to the area under the line.

Midpoint rule and linear functions: For a linear function $f(x) = mx + c$, the function value at the midpoint of $[x_{i-1}, x_i]$ equals $\frac{f(x_{i-1}) + f(x_i)}{2}$, which is the average height. The area of the rectangle $f(\bar{x}_i)\,\Delta x$ then equals the area of the trapezoid under the line. Why the correct answer works: For linear functions, the midpoint value on each subinterval is exactly the average of the endpoint values, so the midpoint rectangle has the same area as the actual region under the line segment. The sum is therefore exact for any $n$. Why distractors fail: Option A is too restrictive — the result holds for all linear functions, not just constants. Option B incorrectly asserts that the midpoint sum is never exact. Option D adds a baseless condition about parity of $n$.

Answer: $\lim_{n \to \infty} \sum_{i=1}^{n} \left(3\left(2 + \frac{4i}{n}\right) - 1\right) \cdot \frac{4}{n}$

Identify the parameters: For $\int_2^6 (3x-1)\,dx$: $a = 2$, $b = 6$, so $\Delta x = \frac{6-2}{n} = \frac{4}{n}$ and $x_i = 2 + \frac{4i}{n}$. Construct the Riemann sum: The right Riemann sum is $\sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} \left(3\left(2 + \frac{4i}{n}\right) - 1\right) \cdot \frac{4}{n}$. Why distractors fail: Option B uses $\Delta x = 6/n$ and starts from $x = 0$ instead of $x = 2$. Option C uses $\Delta x = 6/n$, the wrong interval width. Option D sums from $i = 0$ to $n$, giving $n+1$ terms instead of $n$.

Answer: $11$

Apply linearity of the integral: By the linearity property: $\int_1^4 [2f(x) - 3g(x)]\,dx = 2\int_1^4 f(x)\,dx - 3\int_1^4 g(x)\,dx$. Substitute the given values: $= 2(10) - 3(3) = 20 - 9 = 11$. Why distractors fail: Option A ($17$) may come from $2(10) - 3$. Option C ($23$) could result from adding instead of subtracting: $2(10) + 3(1)$ or a similar error. Option D ($7$) may result from $10 - 3$.

Answer: The left Riemann sum overestimates the integral.

Decreasing functions and left sums: For a strictly decreasing function, the left endpoint of each subinterval gives the maximum value of $f$ on that subinterval. Each rectangle is therefore taller than the curve over the rest of the subinterval. Why the correct answer works: Since $f(x_{i-1}) > f(x)$ for all $x \in (x_{i-1}, x_i]$ on a decreasing function, the left Riemann sum overestimates $\int_0^{10} f(x)\,dx$. Why distractors fail: Option A reverses the relationship (that's true for increasing functions). Option B is almost never true for finite $n$. Option D is incorrect because the monotonicity of $f$ is sufficient to determine the direction of error.

Answer: Left Riemann sum (underestimate) and trapezoidal sum (overestimate)

Effect of monotonicity: Since $e^x$ is strictly increasing, the left Riemann sum underestimates and the right Riemann sum overestimates. Effect of concavity: Since $e^x$ is concave up ($f'' > 0$), the trapezoidal rule overestimates and the midpoint rule underestimates. Combine the analysis: To get an underestimate and an overestimate, the student can use the left Riemann sum (underestimate due to increasing) and the trapezoidal sum (overestimate due to concave up). This is Option C. Why distractors fail: Option A reverses the over/under for left and right sums on an increasing function. Option B reverses the over/under for midpoint and trapezoidal sums on a concave up function. Option D incorrectly labels the trapezoidal sum as an underestimate.

Answer: $1$

Factor and apply the summation formula: $\sum_{i=1}^{n} \frac{2i}{n^2} = \frac{2}{n^2} \sum_{i=1}^{n} i = \frac{2}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{n^2} = \frac{n+1}{n}$. Take the limit: $\lim_{n \to \infty} \frac{n+1}{n} = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = 1$. Why distractors fail: Option A ($0$) would be the limit of $\frac{1}{n}$, not $\frac{n+1}{n}$. Option B ($2$) forgets the cancellation with the factor of 2. Option D ($1/2$) confuses partial simplification steps.

Answer: $5$

Apply additivity over intervals: The property $\int_0^5 f(x)\,dx = \int_0^3 f(x)\,dx + \int_3^5 f(x)\,dx$ allows us to solve for the unknown integral. Solve: $\int_3^5 f(x)\,dx = \int_0^5 f(x)\,dx - \int_0^3 f(x)\,dx = 12 - 7 = 5$. Why distractors fail: Option A ($19$) adds instead of subtracting. Option B ($-5$) reverses the subtraction. Option D ($7$) simply repeats one of the given values.

Answer: The linearity property

Identify the property: The ability to split an integral of a sum into the sum of integrals is part of the linearity property of definite integrals. Linearity also includes the ability to factor out constants: $\int_a^b c\,f(x)\,dx = c\int_a^b f(x)\,dx$. Why the correct answer works: Linearity encompasses both additive splitting of integrands and scalar multiplication, which is exactly what the given equation demonstrates. Why distractors fail: Option A (comparison) relates to inequalities between integrals when $f(x) \le g(x)$. Option B (additivity over intervals) refers to splitting the interval: $\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$. Option D (reversal of limits) refers to $\int_a^b f\,dx = -\int_b^a f\,dx$.

Answer: The trapezoidal estimate is the average of the left and right Riemann sums.

Trapezoidal rule formula: The trapezoidal rule approximates each strip as a trapezoid. Its total value is $T_n = \frac{L_n + R_n}{2}$, the arithmetic mean of the left and right Riemann sums. Why the correct answer works: By construction, each trapezoid uses both endpoint values, which is algebraically equivalent to averaging the left-endpoint rectangle and the right-endpoint rectangle on each subinterval. Why distractors fail: Option A ignores the right sum contribution. Option C is not generally true — for a concave down function, the trapezoidal estimate is less than both. Option D is incorrect; the midpoint and trapezoidal approximations are generally different (though they share the same order of error improvement over left/right sums).

Answer: No, reversing the limits changes the sign: $\int_3^1 f(x)\,dx = -\int_1^3 f(x)\,dx$.

Reversal of limits property: A fundamental property of the definite integral states: $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$. Reversing the limits negates the integral. Why the student's reasoning fails: While the definite integral can represent area, the signed nature of the integral means the direction of integration matters. The integral is not simply an unsigned area. Why distractors fail: Option A and Option D both incorrectly assert that the order of limits doesn't matter. Option C claims doubling, which has no basis in the definition.

Answer: The left Riemann sum underestimates $\int_a^b f(x)\,dx$.

Behavior for increasing functions: For a strictly increasing function, the left endpoint of each subinterval gives the minimum function value on that subinterval, so the left Riemann sum uses heights that are always less than or equal to the actual function values across each subinterval. Why the correct answer works: Since $f$ is increasing, $f(x_{i-1}) < f(x)$ for all $x \in (x_{i-1}, x_i]$, so the left Riemann sum underestimates the true area. Why distractors fail: Option A states the opposite. Option B is incorrect because the right Riemann sum overestimates for increasing functions (it uses the maximum on each subinterval). Option D is only true if $f$ is constant.

Answer: The function value at the left endpoint of each subinterval

Definition of left Riemann sum: A left Riemann sum approximates the area under a curve by constructing rectangles whose heights are determined by the function value at the left endpoint of each subinterval. Why the correct answer works: By definition, the left Riemann sum uses $f(x_{i-1})$ — the left endpoint — as the height of the rectangle on the $i$-th subinterval $[x_{i-1}, x_i]$. Why distractors fail: Option A describes the right Riemann sum. Option C describes the trapezoidal rule (which averages both endpoint values). Option D describes the midpoint Riemann sum.

Answer: $\lim_{n \to \infty} \sum_{i=1}^{n} \left[\left(1 + \frac{3i}{n}\right)^3 + 2\right] \cdot \frac{3}{n}$

Determine the parameters: For $\int_1^4$: $a = 1$, $b = 4$, $b - a = 3$, so $\Delta x = \frac{3}{n}$ and $x_i = 1 + \frac{3i}{n}$. Construct the sum: The right Riemann sum is $\sum_{i=1}^{n} f(x_i)\,\Delta x = \sum_{i=1}^{n} \left[\left(1 + \frac{3i}{n}\right)^3 + 2\right] \cdot \frac{3}{n}$. Why distractors fail: Option B starts from $x = 0$ with $\Delta x = 4/n$, representing $\int_0^4$ instead. Option C sums $n+1$ terms ($i = 0$ to $n$), which is not the standard right Riemann sum. Option D uses $\Delta x = 4/n$, giving an incorrect interval length.

Answer: $\frac{8}{3}$

Set up the Riemann sum: $\Delta x = \frac{2}{n}$, $x_i = \frac{2i}{n}$. The right Riemann sum is $R_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{8}{n^3} \sum_{i=1}^{n} i^2$. Apply the summation formula: $R_n = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{8(n+1)(2n+1)}{6n^2} = \frac{4(n+1)(2n+1)}{3n^2}$. Take the limit: $\lim_{n \to \infty} \frac{4(n+1)(2n+1)}{3n^2} = \frac{4 \cdot 2}{3} = \frac{8}{3}$, since $(n+1)/n \to 1$ and $(2n+1)/n \to 2$. Why distractors fail: Option A ($4/3$) uses $b = 1$ instead of $b = 2$. Option C ($4$) omits the division by 3 from the summation formula. Option D ($2/3$) might result from an error in the $\Delta x$ computation.