📝 Riemann sums explained (24 MCQs)

📖 Explanation: Doubling the number of rectangles generally increases the accuracy of a Riemann sum because the rectangles better match the curve's shape. However, the error doesn't necessarily halve uniformly, especially for non-linear functions or if the function isn't monotonic. Options A and B misunderstand the relationship between rectangle count and area, while D is incorrect.

📖 Explanation: For a constant function, any Riemann sum (left, right, midpoint) is exact, regardless of the number of rectangles. In contrast, functions with high curvature, rapid oscillations, or discontinuities require many more subintervals to achieve a good approximation. The error for a constant function is always zero.

📖 Explanation: For an increasing function like $f(x) = x^3$, the left Riemann sum uses the minimum value on each subinterval. This consistently underestimates the area under the curve, making it an underestimate for all $n$. The approximation improves as $n$ increases, but the nature of the error (underestimation) remains constant.

📖 Explanation: A non-uniform partition with a mesh size of 1 does not guarantee accuracy. For a non-constant function, the error depends on how the subintervals are chosen and the values of $f(x_k^*)$. A uniform partition with a very small mesh size is typically preferred for accurate approximations. The error is not zero unless the function is constant, and it's certainly not infinite for a continuous function.

📖 Explanation: Without knowing the exact values of the sums, we cannot compare them. The right sum with $n=2$ might be larger or smaller than the left sum with $n=4$, depending on how the function behaves on the interval. This question tests the understanding that an increase in $n$ doesn't guarantee the sum is closer to the true value if the method (left vs. right) changes.

📖 Explanation: Since $\sqrt{x}$ is increasing, a right Riemann sum is an overestimate. The true area is 5.33, so an approximation of 7.2 is possible with a moderate number of rectangles. However, the question implies a significant error, suggesting a calculation error rather than a conceptual one. The student might have mis-summed the values or used an incorrect $\Delta x$.

📖 Explanation: A Riemann sum of the velocity function approximates the net signed area, which corresponds to displacement. A negative value for the Riemann sum implies that the particle's displacement (final position minus initial position) is negative over that time interval. The particle may have moved back and forth, but its net change in position is negative.

📖 Explanation: The midpoint rule tends to be more accurate than the trapezoidal rule for functions that are concave down, as the trapezoids overestimate the area while the midpoint rectangles more accurately capture the area under the curve. For concave up functions, the trapezoidal rule is often more accurate.

📖 Explanation: The general form of a Riemann sum is $\sum f(x_k^*) \Delta x$. Here, $\Delta x = \frac{2}{n}$, so the interval is $[0,2]$. The points are $x_k^* = \frac{2k}{n}$. The sum is $\sum_{k=1}^n (1 + x_k^*)^2 \Delta x$, which corresponds to the integral of $(1+x)^2$ from 0 to 2. By substitution, $u = 1+x$, this is equivalent to $\int_1^3 u^2 \, du$.

📖 Explanation: A midpoint rule with a small number of subintervals, like 10, can produce an approximation that is relatively close but not exact. With $n=100$, the approximation would be very close to the true value (error very small). With $n=1$, the error would be very large. The given approximation 52.3 is moderately close to 53.6, suggesting a small to moderate $n$. Thus $n=10$ is the most reasonable.

📖 Explanation: A right Riemann sum overestimates for increasing functions and underestimates for decreasing functions. For $f(x) = -x^2$ on $[0,1]$, the function is decreasing, so the right endpoint is the minimum value on the subinterval, leading to an underestimate. This is a valid counterexample to the claim that right sums always overestimate.

📖 Explanation: For a function that is increasing and concave down, the midpoint rule is generally the most accurate. The left sum underestimates (because the function is increasing), the right sum overestimates, and the trapezoidal rule also overestimates for concave down functions. The midpoint rule tends to balance the errors from the increasing and concave down properties, often yielding the best approximation.

📖 Explanation: In a right Riemann sum, the height of each rectangle (or in this case, the cross-sectional area) is taken from the right endpoint of each subinterval. If the cross-sectional area is increasing along the length of the dam (which is typical), this method will use the largest area in each section, thus overestimating the total volume.

📖 Explanation: The true value is $\int_0^1 x^2 \, dx = \frac{1}{3} \approx 0.3333$. The average of the left and right Riemann sums, $(0.21875 + 0.46875)/2 = 0.34375$, is very close to the true value. In fact, for increasing functions, the trapezoidal rule, which is the average of left and right sums, is a good approximation. Options C and D are common erroneous answers based on misapplying the formula or misreading the problem.

📖 Explanation: The integral $\int_0^\pi \sin(x) \, dx = 2$. A left Riemann sum with $n=100$ gives an approximation very close to 2.0001, indicating the true value is indeed 2. This demonstrates that with a sufficiently large number of subintervals, a left Riemann sum can provide an accurate approximation to the definite integral.

📖 Explanation: The true value of this integral is $\pi/4 \approx 0.785$. The left sum (0.71) underestimates, and the right sum (0.81) overestimates. The trapezoidal rule, which averages the left and right sums, is often a better estimate. The average of 0.71 and 0.81 is 0.76. This is closer to the true value than either endpoint alone.

📖 Explanation: For monotonic functions, the midpoint rule generally lies between the left and right sums. However, this is not true for all functions. For functions that oscillate or have multiple extrema, the midpoint rule can be outside the interval defined by the left and right sums. The key is that the midpoint rule uses the function's value at the center, which might be higher than both endpoints or lower than both.

📖 Explanation: A positive Riemann sum for a signed area indicates that the positive areas (above the x-axis) outweigh the negative areas (below the x-axis). The net signed area is positive. It doesn't mean the student made a mistake or that the function is not integrable; it simply reflects the relative magnitudes of the positive and negative portions of the function over the interval.

📖 Explanation: To demonstrate convergence, one typically starts with a smaller $n$ and then doubles or increases it by a factor of 10 to show the approximation getting closer to the true value. A sequence like $n = 1, 10, 100, 1000, 10000$ clearly shows how the Riemann sum improves as $n$ increases. A more balanced sequence like A is also valid, but B provides a wider range of $n$ values to illustrate convergence.

📖 Explanation: Since $P(t)$ is increasing (exponential growth), its derivative is also increasing. A left Riemann sum uses the minimum value of the derivative on each subinterval, thereby underestimating the total change in population. This is a systematic error due to the choice of method for an increasing function.

📖 Explanation: The definition of the definite integral is the limit of Riemann sums as the mesh size goes to zero (or as $n \to \infty$ for uniform partitions). Therefore, any Riemann sum (left, right, midpoint, etc.) will converge to the same value, which is the definite integral. For $e^x$, this limit is $e - 1$.

📖 Explanation: For a non-linear function, the error in a right Riemann sum depends on the function's behavior. If $\rho(x)$ is increasing, the right sum overestimates; if it's decreasing, it underestimates. The magnitude of the error is related to how much the function curves (its concavity). Using more samples (larger $n$) generally reduces this error.

📖 Explanation: The midpoint rule and the trapezoidal rule are different. The midpoint rule samples the function at the midpoint of each subinterval. The trapezoidal rule averages the left and right endpoint values for each subinterval. While both are approximations, they are distinct methods, and the average of left and right sums defines the trapezoidal rule, not the midpoint rule.

📖 Explanation: For a symmetric function, the overestimation from the increasing part might cancel the underestimation from the decreasing part. However, this cancellation is not guaranteed to be exact, and the error depends on the specific symmetry and the choice of subintervals. The problem is designed to test whether students understand that the error doesn't simply disappear because of symmetry; it depends on the specific form of the function and the partition.