Answer: $38$ litres

Identify the setup: We have $n = 3$ subintervals: $[0,2], [2,4], [4,6]$ with $\Delta t = 2$ minutes. Since $r(t)$ is a rate (litres/min), the integral $\int_0^6 r(t)\,dt$ gives total volume. Compute the left sum: Left endpoints are $t = 0, 2, 4$. $L_3 = \Delta t \cdot [r(0) + r(2) + r(4)] = 2(5 + 8 + 6) = 2 \times 19 = 38$ litres. Why distractors fail: Option A ($34$) is the right sum: $2(8+6+3)=34$. Option C ($44$) results from using all four data values as if there were four subintervals: $2(5+8+6+3)=44$. Option D ($22$) is the sum of left-endpoint function values without multiplying by $\Delta t$.

Answer: The lower and upper limits of integration defining the interval on the $x$-axis

Interpret the notation: In $\int_a^b f(x)\,dx$, the symbol $\int$ denotes integration, $f(x)$ is the integrand, $dx$ indicates the variable of integration, and $a$ and $b$ are the lower and upper limits of integration. Why the correct answer works: Option B correctly identifies $a$ and $b$ as the endpoints of the interval $[a, b]$ over which we integrate — they define where the accumulation begins and ends on the $x$-axis. Why distractors fail: Option A confuses the limits of integration with the range of $f$. Option C confuses the limits with the number of partitions $n$, which appears in a Riemann sum but not in the integral notation. Option D confuses the limits with geometric properties of rectangles in the approximation.

Answer: $14$

Determine $\Delta x$ and partition points: $\Delta x = \frac{4-0}{4} = 1$. The partition points are $x_0=0,\; x_1=1,\; x_2=2,\; x_3=3,\; x_4=4$. Evaluate $f$ at left endpoints: Left endpoints: $x_0=0, x_1=1, x_2=2, x_3=3$. So $f(0)=0$, $f(1)=1$, $f(2)=4$, $f(3)=9$. Compute the sum: $L_4 = \Delta x\,[f(0)+f(1)+f(2)+f(3)] = 1\cdot(0+1+4+9) = 14$. Why distractors fail: Option B ($30$) is the right Riemann sum: $1(1+4+9+16)=30$. Option C ($20$) may result from averaging left and right sums incorrectly. Option D ($21\frac{1}{3}$) is the exact value of $\int_0^4 x^2\,dx = \frac{64}{3}$, not the Riemann sum approximation.

Answer: The student confused signed area (the integral) with total unsigned area, which requires $\int_0^6 |f(x)|\,dx$.

Verify the student's computation: $L_6 = 1 \cdot [f(0)+f(1)+f(2)+f(3)+f(4)+f(5)] = 4+3+1+(-2)+(-3)+(-1) = 2$. The arithmetic is correct. Identify the conceptual error: The value $L_6 = 2$ approximates the signed area $\int_0^6 f(x)\,dx$, where regions below the $x$-axis contribute negatively. The total geometric area between the curve and the $x$-axis is $\int_0^6 |f(x)|\,dx$, which would be approximately $4+3+1+2+3+1 = 14$ using the same left-endpoint values with absolute values. Why distractors fail: Option A is irrelevant — the choice of left vs. right endpoints does not address the signed vs. unsigned area confusion. Option C is about precision, not the conceptual error. Option D is factually wrong; the computation correctly includes the factor $\Delta x = 1$.

Answer: The left sum underestimates and the right sum overestimates the integral.

Identify the behavior of $f$: $f(x) = 3x + 1$ is strictly increasing on $[0, 2]$. Apply the monotonicity rule: For a strictly increasing function, the left-endpoint rectangle on each subinterval has height $f(x_k) < f(x)$ for all $x$ in $(x_k, x_{k+1})$, so the left sum underestimates. Conversely, the right-endpoint rectangle has height $f(x_{k+1})$, which is the maximum value on $[x_k, x_{k+1}]$, so the right sum overestimates. Why distractors fail: Option A reverses the relationship. Option C is false — for a finite $n$, the left and right sums of a non-constant linear function are not equal to the integral (though their average equals the integral). Option D is false — the left and right sums differ for any non-constant function when $n$ is finite.

Answer: $10$

Read function values at right endpoints: With $n = 4$ on $[0, 4]$, $\Delta x = 1$. Right endpoints: $x = 1, 2, 3, 4$. Evaluating $f(x) = -x^2 + 4x$: $f(1) = -1+4 = 3$, $f(2) = -4+8 = 4$, $f(3) = -9+12 = 3$, $f(4) = -16+16 = 0$. Compute the right Riemann sum: $R_4 = 1 \cdot (3 + 4 + 3 + 0) = 10$. Why distractors fail: Option A ($12$) may result from a midpoint calculation or an arithmetic error such as incorrectly evaluating $f(4) = 4$ instead of $0$. Option C ($\frac{32}{3} \approx 10.67$) is the exact value of the integral, not the Riemann sum approximation. Option D ($8$) likely comes from omitting one of the rectangle contributions.

Answer: An approximation of the area under a curve using a finite number of rectangles

Recall the definition: A Riemann sum approximates the area under a curve $f(x)$ on $[a, b]$ by partitioning the interval into $n$ subintervals and summing the areas of rectangles whose heights are determined by function values at chosen sample points. Why the correct answer works: Option B correctly identifies that a Riemann sum uses a finite number of rectangles to approximate (not compute exactly) the area under a curve. Why distractors fail: Option A describes the definite integral (the limit of Riemann sums), not the Riemann sum itself. Option C conflates Riemann sums with the Fundamental Theorem of Calculus. Option D describes the mean value theorem for integrals, which is a different concept.

Answer: Both students are correct; the integral computes signed area.

Recall signed area: The definite integral $\int_a^b f(x)\,dx$ computes the signed area between $f(x)$ and the $x$-axis. Regions where $f(x) \ge 0$ contribute positive area, and regions where $f(x) < 0$ contribute negative area. Evaluate both claims: The first student is correct: when $f(x) \ge 0$, the signed area equals the geometric area. The second student is also correct: when $f(x) < 0$, the integral is negative because every rectangle in the Riemann sum has a negative height. Why distractors fail: Option A is wrong because integrals can be negative. Option B is wrong because the integral does represent area (signed area). Option D is wrong because the sign of $f(x)$ directly determines the sign of the integral.

Answer: Trapezoidal rule with $\Delta t = 3$ and divide by $12$

Assess what data is available: We have function values at the endpoints of each subinterval ($t = 0, 3, 6, 9, 12$) but not at midpoints. The midpoint rule would require $T(1.5), T(4.5), T(7.5), T(10.5)$, which are not given. Choose the most accurate feasible method: The trapezoidal rule uses the same endpoint data as the left and right sums but averages them: $T_n = \frac{\Delta t}{2}[T(0) + 2T(3) + 2T(6) + 2T(9) + T(12)]$. It has $O(\Delta t^2)$ error, compared to $O(\Delta t)$ for left/right sums, making it more accurate. Compute the estimate: $\int_0^{12} T(t)\,dt \approx \frac{3}{2}[72 + 2(78) + 2(85) + 2(80) + 74] = \frac{3}{2}[72 + 156 + 170 + 160 + 74] = \frac{3}{2}(632) = 948$. Average temperature $\approx 948/12 = 79$ °C. Why distractors fail: Options A and B are valid but less accurate for the same data — both have $O(\Delta t)$ error while the trapezoidal rule has $O(\Delta t^2)$ error. Option D uses a single data point and ignores all other information, giving a poor and biased estimate.

Answer: $x_0, x_1, x_2, \ldots, x_{n-1}$ (the left endpoint of each subinterval)

Define the partition: On $[a, b]$ with $n$ equal subintervals, the partition points are $x_k = a + k\Delta x$ for $k = 0, 1, \ldots, n$, where $\Delta x = \frac{b-a}{n}$. Identify the left sum sample points: A left Riemann sum evaluates $f$ at the left endpoint of each subinterval: $L_n = \sum_{k=0}^{n-1} f(x_k)\,\Delta x$. So the sample points are $x_0, x_1, \ldots, x_{n-1}$. Why distractors fail: Option B describes right-endpoint sample points. Option C describes midpoint sample points. Option D describes a general Riemann sum, not specifically the left sum.

Answer: $R_n < I < M_n < L_n$

Use monotonicity for left and right sums: Since $f$ is strictly decreasing, on each subinterval the left endpoint gives the maximum and the right endpoint gives the minimum. Therefore $R_n < I < L_n$. Use concavity for the midpoint sum: For a concave-down function, the midpoint rule overestimates the integral. This is because the chord connecting the endpoints of each subinterval lies below the curve, while the midpoint rectangle captures more area. Formally, for concave-down $f$, $M_n > I$. Combined with the decreasing property: $R_n < I < M_n < L_n$. Why distractors fail: Option A places $M_n$ between $R_n$ and $I$, but $M_n > I$ for concave-down functions. Option C places $M_n < R_n$, which contradicts $M_n > I > R_n$. Option D places $L_n < M_n$, but $L_n$ is the largest overestimate since it uses the maximum value on each subinterval of a decreasing function.

Answer: No — the right sum overestimates because $e^x$ is increasing, not because it is concave up.

Distinguish monotonicity from concavity: For the right Riemann sum, overestimation vs. underestimation is determined by whether $f$ is increasing or decreasing on the interval. If $f$ is increasing, the right endpoint gives the maximum value on each subinterval, so the right sum overestimates. Clarify the role of concavity: Concavity affects the error magnitude and is relevant for rules like the trapezoidal rule and midpoint rule (concave up functions cause the trapezoidal rule to overestimate and the midpoint rule to underestimate). But for left/right sums, monotonicity alone determines the direction of the error. Why distractors fail: Option B incorrectly attributes the overestimation to concavity. Option C contradicts the numerical evidence ($1.7238 > 1.7183$). Option D incorrectly claims both properties are needed — any increasing function (concave up, concave down, or neither) produces a right-sum overestimate.

Answer: Yes — this is the limit of right Riemann sums, which converges to the definite integral for any continuous function.

Parse the limit expression: With $\Delta x = \frac{b-a}{n}$, the sample points $x_k = a + k\Delta x$ for $k = 1, \ldots, n$ are exactly the right endpoints of $n$ equal subintervals of $[a, b]$. Apply the definition of the definite integral: For a continuous function on $[a, b]$, the limit of any Riemann sum (left, right, midpoint, or arbitrary sample points) as $n \to \infty$ with $\|\Delta x\| \to 0$ equals $\int_a^b f(x)\,dx$. The choice of sample point does not matter in the limit. Why distractors fail: Option A is wrong because both left and right sums converge to the integral; there is no requirement to use left endpoints. Option B is wrong because $\frac{b-a}{n}$ is the correct $\Delta x$ for the interval $[a,b]$. Option D is wrong because the definition of the definite integral via Riemann sums applies to all continuous functions, regardless of sign.

Answer: $0$

Observe the symmetry: On $[0,1]$, $\sin(\pi x) \ge 0$, contributing positive area. On $[1,2]$, $\sin(\pi x) \le 0$, contributing negative area of equal magnitude. Apply signed area cancellation: By symmetry, $\int_0^1 \sin(\pi x)\,dx = \frac{2}{\pi}$ and $\int_1^2 \sin(\pi x)\,dx = -\frac{2}{\pi}$. The total is $\frac{2}{\pi} + (-\frac{2}{\pi}) = 0$. Why distractors fail: Option A ($\frac{2}{\pi}$) is only the area of the positive lobe. Option C ($\frac{4}{\pi}$) is the total unsigned area (sum of absolute values of both lobes). Option D ($-\frac{2}{\pi}$) is only the contribution from the negative lobe.

Answer: Evaluating $f$ at the midpoint tends to balance the overestimation and underestimation on each subinterval, reducing overall error.

Understand the error behavior: For a monotonic function on a subinterval, the left and right endpoints give values that are consistently too large or too small. The midpoint, being in the middle, tends to split this difference, producing a rectangle whose area more closely matches the true area under the curve on that subinterval. Why the correct answer works: Option B captures the geometric intuition: the midpoint value averages out the tendency to overestimate or underestimate, leading to cancellation of errors. Why distractors fail: Option A is false — all three methods use the same number $n$ of rectangles. Option C is false — the midpoint rule is still an approximation for most functions. Option D is too restrictive; the midpoint rule applies to any integrable function, not only linear ones (though it is exact for linear functions).

Answer: Because $x^3$ is an odd function, so the positive area on $[0, 2]$ exactly cancels the negative area on $[-2, 0]$.

Recall odd function symmetry: A function $f$ is odd if $f(-x) = -f(x)$. For $f(x) = x^3$, we have $(-x)^3 = -x^3$, confirming it is odd. Apply the signed area interpretation: For an odd function integrated over a symmetric interval $[-a, a]$, the positive signed area on $[0, a]$ is exactly the negative of the signed area on $[-a, 0]$. These cancel, giving $\int_{-a}^a f(x)\,dx = 0$. Why distractors fail: Option A is false — $x^3$ has the antiderivative $\frac{x^4}{4}$. Option C is false — the left and right sums for cubics are not generally equal. Option D overgeneralizes; symmetric intervals only yield zero integrals for odd functions, not all functions (e.g., $\int_{-2}^{2} x^2\,dx \ne 0$).