Riemann Sums & Notation - Calculus 2

1. Which of the following summation expressions, when the limit is taken as $n \to \infty$, represents $\int_0^1 e^{-2x} dx$?
   Answer: $\sum_{i=1}^n e^{-2i/n} \frac{1}{n}$ For $\int_0^1 e^{-2x} dx$, we have $a=0, b=1, f(x)=e^{-2x}$. The width of subintervals is $\Delta x = (1-0)/n = 1/n$. Using right endpoints, $x_i = a + i\Delta x = 0 + i/n = i/n$. The right Riemann sum is $\sum_{i=1}^n f(x_i) \Delta x = \sum_{i=1}^n e^{-2(i/n)} \frac{1}{n}$. This matches choice C. Choice A uses $\Delta x = 2/n$, which would correspond to an interval of length 2. Choice B uses the function $e^{-x}$. Choice D uses right endpoints $x_{i+1}$ but sums from $i=0$ which is the same as a standard right-hand sum, but has $\Delta x = 2/n$.
2. The expression $\sum_{j=1}^{4} \frac{1}{2} \cdot h\left(1 + \frac{j}{2}\right)$ represents a Riemann sum approximation. What interval and which endpoints (left, right, or midpoint) are being used?
   Answer: Interval $[1, 3]$ using right endpoints with 4 equal subintervals When analyzing Riemann sums, you need to identify three key components: the interval being approximated, the width of each subinterval ($\Delta x$), and which point within each subinterval is being used for the function evaluation. Let's decode this expression step by step. The factor $\frac{1}{2}$ represents $\Delta x$, so each subinterval has width $\frac{1}{2}$. With 4 subintervals (since $j$ runs from 1 to 4), the total interval length is $4 \times \frac{1}{2} = 2$. The function evaluations occur at points $1 + \frac{j}{2}$ where $j = 1, 2, 3, 4$. This gives us evaluation points at $1.5, 2, 2.5,$ and $3$. These are the endpoints of our subintervals. Since we're using 4 subintervals of width $\frac{1}{2}$ starting from some point, our subintervals are: $[1, 1.5], [1.5, 2], [2, 2.5], [2.5, 3]$. The evaluation points $1.5, 2, 2.5, 3$ are the right endpoints of these intervals, confirming we're using right endpoints on interval $[1, 3]$. Choice A incorrectly identifies left endpoints. If we used left endpoints, we'd evaluate at $1, 1.5, 2, 2.5$, not $1.5, 2, 2.5, 3$. Choice C misidentifies the interval entirely. Choice D suggests midpoints, but midpoints of $[1, 1.5], [1.5, 2], [2, 2.5], [2.5, 3]$ would be $1.25, 1.75, 2.25, 2.75$. Study tip: Always work backwards from the evaluation points to determine the subintervals, then check whether your points fall at the left endpoints, right endpoints, or midpoints of those intervals.
3. For a continuous function $f(x)$ on $[a, b]$, the finite sum $S = \sum_{i=1}^{n} f(x_{i-1}) \Delta x$ is known as the left-hand Riemann sum. Which change to this expression would convert it to the corresponding right-hand Riemann sum, $R_n$?
   Answer: Change the index of the sample point from $x_{i-1}$ to $x_i$. When you encounter Riemann sum questions, focus on understanding how the sample points are chosen within each subinterval. The key difference between left-hand and right-hand Riemann sums lies in where you evaluate the function within each rectangle. In a left-hand Riemann sum $S = \sum_{i=1}^{n} f(x_{i-1}) \Delta x$, you're using the left endpoint of each subinterval as your sample point. The notation $x_{i-1}$ indicates you're taking the function value at the beginning of the $i$-th subinterval. To convert this to a right-hand Riemann sum, you simply need to evaluate the function at the right endpoint of each subinterval instead, which is represented by $x_i$. This gives you $R_n = \sum_{i=1}^{n} f(x_i) \Delta x$. Answer choice A is incorrect because $x_{i+1}$ would represent a point outside your partition—this doesn't correspond to any standard Riemann sum. Choice B changes the summation limits, which would alter the number of rectangles and create indexing problems since $x_{-1}$ isn't defined in your partition. Choice C halves the width of each rectangle by changing $\Delta x$, which would fundamentally change the approximation method rather than just shifting from left to right endpoints. Remember this pattern: left-hand sums use $x_{i-1}$ (previous endpoint), right-hand sums use $x_i$ (current endpoint), and midpoint sums use the average of these. The summation limits and $\Delta x = \frac{b-a}{n}$ stay the same—only the sample point location changes.
4. Let $f(x)$ be a strictly decreasing and concave up function on the interval $[a, b]$. Let $R_n, L_n, M_n, T_n$ be the right-hand, left-hand, midpoint, and trapezoidal Riemann sum approximations with $n$ subintervals of equal width. Which of the following inequalities must be true?
   Answer: $R_n < M_n < \int_a^b f(x) dx < T_n < L_n$ For a strictly decreasing function, the left-hand sum $L_n$ is an overestimate and the right-hand sum $R_n$ is an underestimate. Thus, $R_n < \int_a^b f(x) dx < L_n$. For a concave up function, the midpoint sum $M_n$ is an underestimate and the trapezoidal sum $T_n$ is an overestimate. Thus, $M_n < \int_a^b f(x) dx < T_n$. Combining these, we know that $R_n$ and $M_n$ are underestimates, while $T_n$ and $L_n$ are overestimates. A more detailed theorem states that for a function that is both decreasing and concave up, the approximations are ordered as follows: $R_n < M_n < \int_a^b f(x) dx < T_n < L_n$.
5. Let $S_n = \sum_{i=1}^{n} (ci + d)$, where $c$ and $d$ are non-zero constants. Which of the following expressions is equivalent to $S_n$?
   Answer: $\frac{n(cn+c+2d)}{2}$ We use the properties of summation to split the expression: $S_n = \sum_{i=1}^{n} ci + \sum_{i=1}^{n} d$. Then we pull out the constants: $S_n = c\sum_{i=1}^{n} i + d\sum_{i=1}^{n} 1$. Using the standard formulas for sums, $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^{n} 1 = n$. Substituting these in gives $S_n = c\frac{n(n+1)}{2} + dn$. To combine these into a single fraction, we find a common denominator: $S_n = \frac{cn(n+1)}{2} + \frac{2dn}{2} = \frac{cn^2 + cn + 2dn}{2}$. Factoring out $n$ from the numerator gives $S_n = \frac{n(cn+c+2d)}{2}$.
6. The velocity of a particle is given by $v(t) = 1/(1+t)$ for $t \ge 0$. Using a left Riemann sum with three equal subintervals on $[0, 3]$, what is the approximate displacement of the particle?
   Answer: 11/6 When you encounter a problem asking for displacement given velocity, you're working with the fundamental theorem of calculus: displacement equals the integral of velocity over time. Since we can't easily integrate $v(t) = \frac{1}{1+t}$ by hand on an exam, we approximate using Riemann sums. For a left Riemann sum on $[0,3]$ with three equal subintervals, each subinterval has width $\Delta t = \frac{3-0}{3} = 1$. The subintervals are $[0,1]$, $[1,2]$, and $[2,3]$. For left Riemann sums, we evaluate the function at the left endpoint of each subinterval. The left endpoints are $t = 0, 1, 2$, giving us: - $v(0) = \frac{1}{1+0} = 1$ - $v(1) = \frac{1}{1+1} = \frac{1}{2}$ - $v(2) = \frac{1}{1+2} = \frac{1}{3}$ The left Riemann sum is: $\Delta t \cdot [v(0) + v(1) + v(2)] = 1 \cdot \left[1 + \frac{1}{2} + \frac{1}{3}\right]$ Converting to a common denominator: $1 + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6}$ Therefore, the approximate displacement is $\frac{11}{6}$, which is answer choice D. Choice A ($\frac{7}{6}$) likely comes from incorrectly adding $1 + \frac{1}{2}$ and forgetting the third term. Choice B ($\frac{13}{12}$) suggests computational errors in finding common denominators. Choice C ($\frac{11}{2}$) appears to be $\frac{11}{6}$ with the denominator incorrectly simplified. Remember: for left Riemann sums, always use the left endpoint of each subinterval, and double-check your fraction arithmetic—it's a common source of errors.
7. Let $S_n = \sum_{i=1}^n (\frac{2i}{n^2} - \frac{3}{n})$. What is the value of $\lim_{n\to\infty} S_n$?
   Answer: -2 We can evaluate this limit by splitting the sum and using summation formulas, or by converting to integrals. Using summation formulas: $S_n = \frac{2}{n^2} \sum_{i=1}^n i - \frac{1}{n} \sum_{i=1}^n 3$. This becomes $S_n = \frac{2}{n^2} \frac{n(n+1)}{2} - \frac{1}{n}(3n) = \frac{n+1}{n} - 3 = 1 + \frac{1}{n} - 3 = -2 + \frac{1}{n}$. Taking the limit, $\lim_{n\to\infty} (-2 + \frac{1}{n}) = -2$. Alternatively, $\lim \sum \frac{2i}{n^2} = \lim \sum 2(\frac{i}{n})\frac{1}{n} = \int_0^1 2x dx = [x^2]_0^1 = 1$. And $\lim \sum \frac{-3}{n} = \int_0^1 -3 dx = [-3x]_0^1 = -3$. The sum of the integrals is $1 + (-3) = -2$.
8. Evaluate the geometric sum $\sum_{k=0}^{n-1} a r^k$. This formula is used in the derivation of some Riemann sums. What is the correct closed-form expression?
   Answer: $a \frac{1-r^{n}}{1-r}$ This is the standard formula for the sum of the first $n$ terms of a geometric series with first term $a$ and common ratio $r$. The sum $S = a + ar + ar^2 + ... + ar^{n-1}$. Multiplying by $r$ gives $rS = ar + ar^2 + ... + ar^n$. Subtracting the two equations gives $S - rS = a - ar^n$, so $S(1-r) = a(1-r^n)$. Dividing by $(1-r)$ yields $S = a \frac{1-r^n}{1-r}$. Distractor A uses $n-1$ in the exponent, which would be correct if the sum went to $n-2$. Distractor B uses $n+1$, which applies if the sum goes to $n$.
9. If $\int_1^5 g(x) \, dx = 12$ and we approximate this integral using a right Riemann sum with $n = 4$ equal subintervals, what is the relationship between the exact value and our approximation if $g(x)$ is strictly decreasing on $[1, 5]$?
   Answer: The right Riemann sum underestimates the integral by some unknown amount For a strictly decreasing function, the right endpoint of each subinterval gives the minimum value of the function on that subinterval. Therefore, the right Riemann sum will underestimate the true area under the curve. We cannot determine the exact amount of underestimation without knowing the specific function g(x). Choice A incorrectly assumes we can calculate the exact error. Choice B has the wrong direction (overestimate vs underestimate). Choice D ignores that Riemann sums are approximations.
10. If $f(x) = x^3$ and we want $\lim_{n \to \infty} \sum_{i=1}^{n} f\left(x_i^*\right) \Delta x = \int_0^2 x^3 \, dx$, which of the following correctly expresses this limit using right endpoints?
    Answer: $\lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^3 \cdot \frac{2}{n}$ This question tests your understanding of Riemann sums and how they connect to definite integrals. When you see a limit of a sum that equals a definite integral, you need to identify the correct partition width and sample points for the given interval. To set up a Riemann sum for $\int_0^2 x^3 \, dx$ using right endpoints, you need two key components. First, the partition width: since you're dividing the interval $[0,2]$ into $n$ equal subintervals, $\Delta x = \frac{2-0}{n} = \frac{2}{n}$. Second, the right endpoints: for subinterval $i$, the right endpoint is $x_i^* = 0 + i \cdot \frac{2}{n} = \frac{2i}{n}$ where $i$ runs from 1 to $n$. The correct expression is therefore $\lim_{n \to \infty} \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^3 \cdot \frac{2}{n}$, which is answer D. Looking at the wrong answers: A uses the correct structure but assumes the interval is $[0,1]$ instead of $[0,2]$ — notice both the sample points and $\Delta x$ are missing the factor of 2. B uses the correct interval setup but runs the index from 0 to $n-1$, which gives left endpoints instead of right endpoints. C attempts right endpoints but incorrectly uses $\frac{2(i-1)}{n}$, which actually gives left endpoints when $i$ runs from 1 to $n$. Study tip: Always check both the interval length (affects $\Delta x$) and endpoint type (affects the sample point formula). Right endpoints with index $i = 1$ to $n$ use $x_i^* = a + i \cdot \Delta x$.
11. If $\lim_{n\to\infty} \sum_{i=1}^n \frac{b}{n} \left(\frac{bi}{n}\right)^2 = 9$, where $b > 0$, what is the value of $b$?
    Answer: 3 The given limit is a right Riemann sum. Let's identify the components. Let $\Delta x = b/n$. Let the sample points be $x_i = bi/n$. This corresponds to an integral over the interval $[0, b]$. The function being summed is $f(x_i) = (x_i)^2$, so $f(x) = x^2$. The limit is therefore equal to the definite integral $\int_0^b x^2 dx$. We are given that this integral equals 9. We evaluate the integral: $\int_0^b x^2 dx = [\frac{x^3}{3}]_0^b = \frac{b^3}{3} - 0 = \frac{b^3}{3}$. Setting this equal to 9, we have $\frac{b^3}{3} = 9$, which means $b^3 = 27$. Since $b$ is positive, we take the real cube root to find $b = 3$.
12. The midpoint Riemann sum $M_n$ is used to approximate $\int_0^\pi \sin(x) dx$. Which expression represents $M_4$?
    Answer: $\frac{\pi}{4} \left( \sin(\frac{\pi}{8}) + \sin(\frac{3\pi}{8}) + \sin(\frac{5\pi}{8}) + \sin(\frac{7\pi}{8}) \right)$ To find the midpoint sum $M_4$ for $\int_0^\pi \sin(x) dx$, we first find $\Delta x = \frac{b-a}{n} = \frac{\pi - 0}{4} = \frac{\pi}{4}$. The four subintervals are $[0, \pi/4], [\pi/4, \pi/2], [\pi/2, 3\pi/4], [3\pi/4, \pi]$. The midpoints of these subintervals are $x_1^* = \pi/8, x_2^* = 3\pi/8, x_3^* = 5\pi/8, x_4^* = 7\pi/8$. The midpoint sum is $M_4 = \sum_{i=1}^4 f(x_i^*) \Delta x = \frac{\pi}{4} (\sin(\frac{\pi}{8}) + \sin(\frac{3\pi}{8}) + \sin(\frac{5\pi}{8}) + \sin(\frac{7\pi}{8}))$. Choice B represents the left sum, Choice C represents the right sum, and Choice D is the midpoint sum for $n=2$.
13. If $\int_a^b f(x) dx = L > 0$, which of the following statements about the Riemann sum $S_n = \sum_{i=1}^n f(x_i^*) \Delta x$ must be true?
    Answer: There exists an integer $N$ such that for all $n > N$, $S_n > 0$. The definite integral is defined as the limit of the Riemann sum: $\lim_{n \to \infty} S_n = L$. The definition of a limit states that for any $\epsilon > 0$, there exists an integer $N$ such that for all $n > N$, $|S_n - L| < \epsilon$. Since we are given $L > 0$, we can choose $\epsilon = L/2$. Then, for $n > N$, we have $-L/2 < S_n - L < L/2$, which implies $L/2 < S_n < 3L/2$. Since $L > 0$, $L/2 > 0$, so $S_n > 0$ for all sufficiently large $n$. Choice A is incorrect because for small $n$, the approximation $S_n$ could be negative. Choice B is incorrect because the net area can be positive even if the function is negative on some parts of the interval. Choice D is unrelated to the value of the integral.
14. The expression $\frac{1}{n} \sum_{k=1}^n \sqrt{1 - (k/n)^2}$ is a Riemann sum for a certain function on the interval $[0, 1]$. The exact value of the limit of this sum as $n \to \infty$ corresponds to the area of which geometric shape?
    Answer: A quarter-circle with radius 1 The expression is a right Riemann sum for the integral $\int_0^1 \sqrt{1-x^2} dx$. The function $y = \sqrt{1-x^2}$ is the equation of the top half of a circle centered at the origin with radius 1 (since $y^2 = 1-x^2 \Rightarrow x^2+y^2=1$). The interval of integration is from $x=0$ to $x=1$. This describes the area under the curve in the first quadrant, which is a quarter of the full circle's area. Therefore, the integral represents the area of a quarter-circle with radius 1.
15. A right Riemann sum is used to approximate $\int_{-2}^{2} x^3 dx$ with $n=4$ subintervals. What is the value of this approximation?
    Answer: 8 For the integral $\int_{-2}^{2} x^3 dx$ with $n=4$, the width of each subinterval is $\Delta x = \frac{2 - (-2)}{4} = 1$. The subintervals are $[-2, -1], [-1, 0], [0, 1], [1, 2]$. For a right Riemann sum, we use the right endpoints of these intervals: $x_1=-1, x_2=0, x_3=1, x_4=2$. The approximation is $R_4 = \sum_{i=1}^4 f(x_i) \Delta x = [f(-1) + f(0) + f(1) + f(2)] \cdot 1$. With $f(x) = x^3$, this is $((-1)^3 + 0^3 + 1^3 + 2^3) \cdot 1 = (-1 + 0 + 1 + 8) \cdot 1 = 8$. Distractor B (0) is the exact value of the integral, a common trap. Distractor C (-8) is the value of the left Riemann sum.
16. Let $f(x) = x^2+1$. The interval $[0, 4]$ is partitioned into $n=4$ subintervals of equal width. What is the difference between the upper sum (circumscribed rectangles) and the lower sum (inscribed rectangles) for this partition?
    Answer: 16 When you encounter Riemann sums with upper and lower approximations, you're exploring how rectangles can bound the area under a curve. The key insight is that for increasing functions, the upper sum uses right endpoints (overestimating area) while the lower sum uses left endpoints (underestimating area). Given $f(x) = x^2 + 1$ on $[0,4]$ with $n=4$ subintervals, each subinterval has width $\Delta x = \frac{4-0}{4} = 1$. The partition points are $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4$. Since $f(x) = x^2 + 1$ is increasing on $[0,4]$, the maximum value on each subinterval occurs at the right endpoint, and the minimum at the left endpoint. Upper sum: $U = 1 \cdot [f(1) + f(2) + f(3) + f(4)] = 1 \cdot [2 + 5 + 10 + 17] = 34$ Lower sum: $L = 1 \cdot [f(0) + f(1) + f(2) + f(3)] = 1 \cdot [1 + 2 + 5 + 10] = 18$ The difference is $U - L = 34 - 18 = 16$, confirming answer D. Looking at the wrong answers: A) 32 likely comes from doubling the correct difference. B) 8 might result from using the wrong width or miscounting terms. C) 4 could arise from confusing the number of subintervals with the actual difference. Remember: for increasing functions, the upper-lower difference always equals $\Delta x \cdot [f(b) - f(a)]$, where $[a,b]$ is your interval. This gives you a quick check: $1 \cdot [17 - 1] = 16$.
17. Consider the limit $\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \sqrt{1 + \left(\frac{2i}{n}\right)^2}$. This limit represents a definite integral. What are the correct limits of integration and integrand?
    Answer: $\int_0^2 \sqrt{1 + x^2} \, dx$ In this Riemann sum, Δx = 2/n, so the interval length is 2. The expression 2i/n represents the right endpoint formula a + iΔx. Since the smallest value (when i=1) is 2/n ≈ 0 and the largest (when i=n) is 2, we have the interval [0,2]. The integrand is √(1 + x²) where x = 2i/n. Choice B has wrong limits. Choice C incorrectly expands (2i/n)² as 4x² instead of x². Choice D has wrong lower limit.
18. A student writes the Riemann sum $\sum_{k=0}^{n-1} f\left(-1 + \frac{3k}{n}\right) \cdot \frac{3}{n}$ and claims it approximates $\int_{-1}^{2} f(x) \, dx$ using left endpoints. Which part of the student's work contains an error?
    Answer: The sum correctly represents a left Riemann sum for $\int_{-1}^{2} f(x) \, dx$ Let's verify: Δx = 3/n, so interval length = 3. The expression -1 + 3k/n gives left endpoints starting at -1 (when k=0) and ending at -1 + 3(n-1)/n = -1 + 3 - 3/n = 2 - 3/n. As n→∞, this approaches 2, confirming the interval is [-1,2]. The sum uses k from 0 to n-1, giving n terms total, which is correct for n subintervals. This is indeed a proper left Riemann sum for ∫₋₁² f(x)dx. The other choices identify non-existent errors.
19. What is the exact value of $\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}\left(1 + \frac{i}{n}\right)$?
    Answer: 3 This limit can be evaluated in two ways. Method 1: Using summation formulas. $\sum_{i=1}^n \frac{2}{n}(1 + \frac{i}{n}) = \frac{2}{n}\sum_{i=1}^n 1 + \frac{2}{n^2}\sum_{i=1}^n i = \frac{2}{n}(n) + \frac{2}{n^2}(\frac{n(n+1)}{2}) = 2 + \frac{n+1}{n} = 2 + 1 + \frac{1}{n} = 3 + \frac{1}{n}$. The limit as $n \to \infty$ is 3. Method 2: Converting to a definite integral. The sum can be written as $2 \lim_{n\to\infty} \sum_{i=1}^n (1 + \frac{i}{n})\frac{1}{n}$. This is $2$ times the right Riemann sum for $f(x) = 1+x$ on $[0,1]$. So the value is $2 \int_0^1 (1+x) dx = 2[x + \frac{x^2}{2}]_0^1 = 2(1 + \frac{1}{2}) = 3$.
20. Let $\int_a^b f(x) dx = L$. Which of the following is equivalent to $\int_a^b (2f(x)+3) dx$ in terms of L, a, and b?
    Answer: $2L + 3b - 3a$ Using the properties of definite integrals, we can split the integral: $\int_a^b (2f(x)+3) dx = \int_a^b 2f(x) dx + \int_a^b 3 dx$. The constant can be pulled out of the first integral: $2\int_a^b f(x) dx + \int_a^b 3 dx$. We are given that $\int_a^b f(x) dx = L$. The second integral evaluates to $[3x]_a^b = 3b - 3a = 3(b-a)$. Substituting these back gives $2L + 3(b-a)$.

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