Riemann Sums & Notation - Calculus 1

1. Which expression represents $\int_0^4 \sqrt{x} \, dx$ as a limit of Riemann sums using right endpoints?
   Answer: $\lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{\frac{4i}{n}} \cdot \frac{4}{n}$ When you encounter a question asking for a Riemann sum representation of a definite integral, you need to carefully construct the sum using the proper formula: $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \cdot \Delta x$, where $\Delta x = \frac{b-a}{n}$ and $x_i$ represents the sample points. For $\int_0^4 \sqrt{x} \, dx$ with right endpoints, start by finding $\Delta x = \frac{4-0}{n} = \frac{4}{n}$. When using right endpoints on the interval $[0,4]$, the sample points are $x_i = 0 + i \cdot \frac{4}{n} = \frac{4i}{n}$ for $i = 1, 2, 3, \ldots, n$. This gives us $f(x_i) = \sqrt{\frac{4i}{n}}$. Therefore, the correct Riemann sum is $\lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{\frac{4i}{n}} \cdot \frac{4}{n}$, which is choice B. Choice A uses the wrong indexing ($i=0$ to $n-1$), which would give you left endpoints instead of right endpoints. Choice C also uses incorrect indexing with $\sqrt{\frac{4(i-1)}{n}}$, effectively shifting all sample points left and again producing left endpoints. Choice D has the correct function evaluation and indexing but uses $\Delta x = \frac{1}{n}$ instead of $\frac{4}{n}$, which would be appropriate for an interval of length 1, not length 4. Remember: for right endpoints on $[a,b]$, always use $x_i = a + i \cdot \frac{b-a}{n}$ with $i$ running from 1 to $n$, and don't forget that $\Delta x = \frac{b-a}{n}$.
2. Let $f$ be a function that is strictly decreasing and concave up on the interval $[a, b]$. Let $L_n, R_n, M_n, T_n$ be the left-hand, right-hand, midpoint, and trapezoidal rule approximations, respectively, for $\int_a^b f(x) \,dx$ with $n$ subintervals. Which of the following inequalities must be true?
   Answer: $T_n > \int_a^b f(x) \,dx$ For a function that is strictly decreasing on $[a, b]$, the left Riemann sum $L_n$ is an overestimate and the right Riemann sum $R_n$ is an underestimate. Thus, $R_n < \int_a^b f(x) \,dx < L_n$. This eliminates choices A and B. For a function that is concave up on $[a, b]$, the trapezoidal rule $T_n$ is an overestimate and the midpoint rule $M_n$ is an underestimate. Thus, $M_n < \int_a^b f(x) \,dx < T_n$. This eliminates choice C and confirms choice D is correct. The information that the function is decreasing is used to eliminate A and B, while the information about concavity is used for C and D.
3. An approximation of $\int_{0}^{4} (x^2 + 1) dx$ is made using a right Riemann sum with $n$ subintervals of equal width, denoted by $S_n = \sum_{i=1}^{n} f(x_i) \Delta x$. Which of the following expressions is equivalent to $S_n$?
   Answer: $\frac{64}{n^3} \sum_{i=1}^{n} i^2 + 4$ For the integral $\int_{0}^{4} (x^2 + 1) dx$ with $n$ subintervals, $\Delta x = \frac{4-0}{n} = \frac{4}{n}$. For a right Riemann sum, the sample points are $x_i = a + i\Delta x = 0 + i\frac{4}{n} = \frac{4i}{n}$. The function is $f(x) = x^2 + 1$, so $f(x_i) = \left(\frac{4i}{n}\right)^2 + 1 = \frac{16i^2}{n^2} + 1$. The Riemann sum is $S_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{16i^2}{n^2} + 1\right) \frac{4}{n}$. Using properties of summation, we can distribute: $S_n = \frac{4}{n} \sum_{i=1}^{n} \left(\frac{16i^2}{n^2} + 1\right) = \frac{4}{n} \left( \sum_{i=1}^{n} \frac{16i^2}{n^2} + \sum_{i=1}^{n} 1 \right)$. We can pull out constants from the sums: $S_n = \frac{4}{n} \left( \frac{16}{n^2} \sum_{i=1}^{n} i^2 + n \right)$. Distributing the $\frac{4}{n}$ gives $S_n = \frac{64}{n^3} \sum_{i=1}^{n} i^2 + \frac{4}{n} \cdot n = \frac{64}{n^3} \sum_{i=1}^{n} i^2 + 4$.
4. The midpoint Riemann sum for $\int_{-1}^{3} x^3 \, dx$ using 4 subintervals has sample points at which $x$-values?
   Answer: $x = -0.5, 0.5, 1.5, 2.5$ When you encounter a midpoint Riemann sum problem, you need to systematically find where the sample points fall within each subinterval. The key is understanding that midpoint means exactly halfway between the left and right endpoints of each subinterval. For this integral from $x = -1$ to $x = 3$ with 4 subintervals, start by finding the width: $\Delta x = \frac{3 - (-1)}{4} = 1$. This gives you four subintervals: $[-1, 0],[0, 1],[1, 2]$, and $[2, 3]$. The midpoint of each subinterval is found by averaging its endpoints. For $[-1, 0]$: $\frac{-1 + 0}{2} = -0.5$. For $[0, 1]$: $\frac{0 + 1}{2} = 0.5$. For $[1, 2]$: $\frac{1 + 2}{2} = 1.5$. For $[2, 3]$: $\frac{2 + 3}{2} = 2.5$. This confirms answer C. Choice A gives you $x = 0, 1, 2, 3$, which are the right endpoints of each subinterval—this would be a right Riemann sum. Choice B gives you $x = -1, 0, 1, 2$, which are the left endpoints—this would be a left Riemann sum. Choice D mixes some correct midpoints ($-0.5$ and 2.5) with incorrect values, representing a common error where students confuse different Riemann sum types. Remember: midpoint Riemann sums always use sample points that are exactly halfway between subinterval boundaries. Calculate $\Delta x$ first, then find the midpoint of each subinterval by averaging its endpoints.
5. The definite integral $\int_2^5 x^3 \,dx$ can be represented as the limit of a Riemann sum. Which of the following expressions does NOT represent this definite integral?
   Answer: $\lim_{n \to \infty} \sum_{i=1}^n \frac{3}{n} \left(\frac{3i}{n}\right)^3$ For the integral $\int_2^5 x^3 \,dx$, we have $a=2, b=5$, and $f(x)=x^3$. The width of each subinterval is $\Delta x = \frac{5-2}{n} = \frac{3}{n}$. The sample points are $x_i = a + i\Delta x = 2 + \frac{3i}{n}$. A) This is the standard right Riemann sum, as $i$ goes from 1 to $n$. B) This is the standard left Riemann sum, as $i$ goes from 0 to $n-1$. C) This is the standard midpoint Riemann sum, using sample points $x_i^* = a + (i - 1/2)\Delta x$. D) This expression represents $\int_0^3 x^3 \,dx$, because the sample points are $x_i = \frac{3i}{n}$, which corresponds to an interval starting at $a=0$ with width $b-a=3$. This does not represent the given integral.
6. The expression $S_n = \frac{2}{n} \sum_{i=1}^{n} \left|1 - \frac{2(i-1)}{n}\right|$ is a Riemann sum approximation for a certain definite integral. What is the exact value of $\lim_{n\to\infty} S_n$?
   Answer: $1$ The expression is a Riemann sum. We can identify $\Delta x = \frac{2}{n}$, which suggests an interval of length 2. The term $x_i^* = \frac{2(i-1)}{n}$ represents the sample points. Since the index is $i-1$, this is a left Riemann sum. With $x_i^* = a + (i-1)\Delta x$, we have $a=0$. The interval is $[0, 2]$. The function is $f(x) = |1-x|$. Therefore, the limit is equal to the definite integral $\int_{0}^{2} |1-x| dx$. To evaluate this integral, we split it at $x=1$, where the expression inside the absolute value changes sign: $\int_{0}^{2} |1-x| dx = \int_{0}^{1} (1-x) dx + \int_{1}^{2} -(1-x) dx = \int_{0}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx$. Geometrically, this is the sum of the areas of two triangles. The first is a triangle with base 1 and height 1 (area 1/2), and the second is also a triangle with base 1 and height 1 (area 1/2). The total area is $\frac{1}{2} + \frac{1}{2} = 1$.
7. The velocity of a particle is given by $v(t)$ for $t \ge 0$. The expression $\sum_{k=1}^{10} v(0.5k) \cdot (0.5)$ represents an approximation of the particle's displacement. Which of the following does this sum represent?
   Answer: A right Riemann sum on the interval $[0, 5]$ with 10 subintervals. Let's analyze the given sum: $\sum_{k=1}^{10} v(0.5k) \cdot (0.5)$. This is a Riemann sum with $n=10$ terms. The width of each subinterval is $\Delta x = 0.5$. The summation index $k$ runs from 1 to 10. The sample points are $t_k = 0.5k$. When $k=1, t_1=0.5$. When $k=10, t_{10}=5$. Since the index starts at 1 and the sample points are $a+k\Delta x$ (with $a=0$), this is a right Riemann sum. The interval ends at $b = a + n\Delta x = 0 + 10(0.5) = 5$. So, the sum represents a right Riemann sum approximation for $\int_0^5 v(t)\,dt$ with 10 subintervals.
8. The limit $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^4$ represents a definite integral. What is the value of this integral?
   Answer: $\frac{1}{5}$ This limit is the definition of a definite integral. We identify the components: $\Delta x = \frac{1}{n}$ and the right endpoints are $x_k = a + k\Delta x$. Since the term inside the function is $\frac{k}{n}$, we can set $a=0$. Then $x_k = \frac{k}{n}$. The interval is $[0, 1]$ since $b-a = n\Delta x = n(\frac{1}{n})=1$. The function is $f(x_k) = (\frac{k}{n})^4$, so $f(x) = x^4$. The limit is equivalent to the integral $\int_0^1 x^4 \,dx$. Evaluating this gives $\left[\frac{x^5}{5}\right]_0^1 = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5}$.
9. Let $S_n = \sum_{i=1}^n f(c_i) \Delta x$ be a Riemann sum for a continuous function $f$ on $[a, b]$ with $\Delta x = (b-a)/n$ and $c_i \in [x_{i-1}, x_i]$. Which of the following statements is not necessarily true?
   Answer: If $f$ is strictly increasing, then the sequence of left Riemann sums $L_n$ is strictly increasing, i.e., $L_n < L_{n+1}$ for all $n \ge 1$. A) This is the definition of the definite integral for a continuous (and thus integrable) function; it is true regardless of the choice of sample points $c_i$. B) The right Riemann sum is a specific type of Riemann sum, so its limit is also the definite integral. This is true. C) For a strictly increasing function, the minimum value on any subinterval $[x_{i-1}, x_i]$ is at the left endpoint $x_{i-1}$. Therefore, the left Riemann sum $L_n$ will be a strict underestimate of the integral. This is true. D) While $L_n$ converges to the integral from below, the sequence $\{L_n\}$ itself is not guaranteed to be monotonically increasing. Adding more rectangles can sometimes slightly decrease the sum value before it continues to increase toward the limit. This statement is not necessarily true.
10. Let $f(x) = x^2$. Consider the right Riemann sum $R_n$ and the left Riemann sum $L_n$ for $f(x)$ on the interval $[0, 2]$ with $n$ equal subintervals. What is the value of $R_n - L_n$?
    Answer: $\frac{8}{n}$ For $f(x)$ on $[a,b]$, $R_n - L_n = (f(b) - f(a)) \Delta x$. In this case, $a=0, b=2$, and $\Delta x = \frac{2-0}{n} = \frac{2}{n}$. The function is $f(x)=x^2$. So, $R_n - L_n = (f(2) - f(0)) \frac{2}{n} = (2^2 - 0^2) \frac{2}{n} = (4) \frac{2}{n} = \frac{8}{n}$. To see this directly: $R_n = \sum_{i=1}^n f(x_i) \Delta x$ and $L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x$. $R_n - L_n = (f(x_n) - f(x_0))\Delta x$. Since $x_n = b = 2$ and $x_0 = a = 0$, the result follows.
11. Which of the following expressions represents the definite integral $\int_{1}^{3} \ln(x) dx$ as the limit of a right Riemann sum?
    Answer: $\lim_{n\to\infty} \sum_{i=1}^{n} \ln\left(1 + \frac{2i}{n}\right) \frac{2}{n}$ For a definite integral $\int_{a}^{b} f(x) dx$, the limit of a right Riemann sum is given by $\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \Delta x$. For the integral $\int_{1}^{3} \ln(x) dx$, we have $a=1$, $b=3$, and $f(x)=\ln(x)$. The width of each subinterval is $\Delta x = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}$. The right endpoints are given by $x_i = a + i\Delta x = 1 + i\left(\frac{2}{n}\right) = 1 + \frac{2i}{n}$. Therefore, $f(x_i) = \ln(x_i) = \ln\left(1 + \frac{2i}{n}\right)$. Substituting these into the Riemann sum formula gives $\lim_{n\to\infty} \sum_{i=1}^{n} \ln\left(1 + \frac{2i}{n}\right) \frac{2}{n}$.
12. The expression $\lim_{n \to \infty} \sum_{i=1}^{n} \sin\left(\frac{\pi i}{n}\right) \cdot \frac{\pi}{n}$ equals which definite integral?
    Answer: $\int_0^{\pi} \sin(x) \, dx$ When you see a limit of a Riemann sum like this, you're looking at the formal definition of a definite integral. The key is identifying the interval of integration and the function being integrated. The general form of a Riemann sum is $\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x$, where $\Delta x = \frac{b-a}{n}$ and $x_i = a + i \cdot \Delta x$. Here, $\Delta x = \frac{\pi}{n}$, which tells us the interval length is $b - a = \pi$. Since $x_i = \frac{\pi i}{n} = 0 + i \cdot \frac{\pi}{n}$, we have $a = 0$, making $b = \pi$. The function being evaluated is $f(x) = \sin(x)$. Therefore, this Riemann sum represents $\int_0^{\pi} \sin(x) \, dx$, which is answer D. Let's examine why the other options are incorrect: A) $\int_1^{\pi} \sin(x) \, dx$ has the wrong lower limit. The Riemann sum starts at $x_1 = \frac{\pi}{n}$, which approaches 0 as $n \to \infty$, not 1. B) $\int_0^{1} \sin(\pi x) \, dx$ has the wrong interval (0 to 1 instead of 0 to π) and wrong function ($\sin(\pi x)$ instead of $\sin(x)$). C) $\int_0^{\pi} \sin(\pi x) \, dx$ has the correct interval but wrong function. The argument inside the sine function in the sum is $\frac{\pi i}{n}$, not $\pi \cdot \frac{\pi i}{n}$. Study tip: When converting Riemann sums to integrals, always identify $\Delta x$ first to find your interval, then determine what function is being evaluated at each $x_i$.
13. Let $L_n$ be the left Riemann sum, $R_n$ be the right Riemann sum, and $M_n$ be the midpoint Riemann sum for $\int_0^2 (x^2-2x) \,dx$ with $n$ equal subintervals. Which statement is true for any $n \ge 2$?
    Answer: $L_n = R_n$. The function $f(x) = x^2-2x = x(x-2)$ is a parabola opening upwards with roots at $x=0$ and $x=2$. This function is symmetric about its axis of symmetry at $x = \frac{0+2}{2} = 1$. Over the interval $[0, 2]$, the function is decreasing on $[0, 1]$ and increasing on $[1, 2]$. Because the function is not monotonic over the entire interval $[0, 2]$, we cannot conclude that $L_n$ or $R_n$ is always an overestimate or underestimate. However, there is a special property for the difference: $R_n - L_n = (f(b) - f(a))\Delta x$. Here, $a=0, b=2$. $f(0)=0$ and $f(2)=2^2-2(2)=0$. Thus, $R_n - L_n = (0-0)\frac{2}{n} = 0$, which means $L_n = R_n$ for any $n$.
14. Let $R_n = \sum_{i=1}^{n} \frac{b-a}{n} f\left(a + i \frac{b-a}{n}\right)$ be the right Riemann sum for $\int_a^b f(x)dx$. Which of the following summations is equivalent to $R_n$?
    Answer: $\sum_{j=2}^{n+1} \frac{b-a}{n} f\left(a + (j-1) \frac{b-a}{n}\right)$ The original sum $R_n$ expands to $\Delta x [f(a+\Delta x) + f(a+2\Delta x) + \dots + f(a+n\Delta x)]$. We need to find an equivalent expression by manipulating the index of summation. A) This is the left Riemann sum $L_n$, as the index $k$ runs from 0 to $n-1$. B) Let's use a substitution $i = j-1$. When $j=2$, $i=1$. When $j=n+1$, $i=n$. The summation becomes $\sum_{i=1}^{n} \frac{b-a}{n} f\left(a + i \frac{b-a}{n}\right)$, which is exactly $R_n$. C) This is the left Riemann sum $L_n$ written with the index running from 1 to $n$. D) This sum is missing the last term of $R_n$ since the index stops at $n-1$.
15. Which of the following limits is equal to $\int_1^3 \frac{e^x}{x} \,dx$?
    Answer: $\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n} \frac{e^{1+2i/n}}{1+2i/n}$ For the integral $\int_1^3 \frac{e^x}{x} \,dx$, we identify $a=1, b=3$, and $f(x)=\frac{e^x}{x}$. The width of the subintervals is $\Delta x = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}$. Using a right Riemann sum, the sample points are $x_i = a + i\Delta x = 1 + \frac{2i}{n}$. The limit of the Riemann sum is $\lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x = \lim_{n\to\infty} \sum_{i=1}^n \frac{e^{1+2i/n}}{1+2i/n} \cdot \frac{2}{n}$. This matches option C. Option A incorrectly uses $a=0$. Option B has the wrong interval width $\Delta x=3/n$. Option D has the wrong interval width $\Delta x=1/n$.
16. What is the value of the limit $L = \lim_{n\to\infty} \sum_{i=1}^{n} \sqrt{4 - \left(\frac{2i}{n}\right)^2} \cdot \frac{2}{n}$?
    Answer: $\pi$ The given limit is the definition of a definite integral using a right Riemann sum. The term $\frac{2}{n}$ represents $\Delta x$. Since $\Delta x = \frac{b-a}{n}$, we can infer that the interval length $b-a$ is 2. The term $\frac{2i}{n}$ represents the sample points $x_i = a + i\Delta x$. If we choose the interval to be $[0, 2]$, then $a=0, b=2$, so $x_i = 0 + i(\frac{2}{n}) = \frac{2i}{n}$. The function being evaluated is $f(x_i) = \sqrt{4 - x_i^2}$, so $f(x) = \sqrt{4-x^2}$. The limit is therefore equal to the definite integral $\int_{0}^{2} \sqrt{4-x^2} dx$. This integral represents the area under the curve $y=\sqrt{4-x^2}$ from $x=0$ to $x=2$. The graph of $y=\sqrt{4-x^2}$ is the upper semi-circle of radius 2 centered at the origin. The integral represents the area of a quarter-circle of radius 2. The area is $\frac{1}{4}\pi r^2 = \frac{1}{4}\pi (2)^2 = \pi$.
17. The limit $L = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n+i}$ is equivalent to which definite integral?
    Answer: $\int_{0}^{1} \frac{1}{1+x} dx$ To identify the definite integral, we need to rewrite the sum in the form $\sum f(x_i) \Delta x$. We can factor out $\frac{1}{n}$ from the term in the summand: $\frac{1}{n+i} = \frac{1}{n(1+i/n)} = \frac{1}{n} \cdot \frac{1}{1+i/n}$. The sum becomes $\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{1+i/n} \cdot \frac{1}{n}$. Now it matches the form of a right Riemann sum. We can identify $\Delta x = \frac{1}{n}$. This means $b-a=1$. The sample points are $x_i = a+i\Delta x$. From the term $\frac{1}{1+i/n}$, we can identify $f(x_i) = \frac{1}{1+x_i}$ with $x_i = i/n$. If $x_i = i/n$, then $a=0$ and the interval is $[0, 1]$. The function is $f(x) = \frac{1}{1+x}$. Therefore, the limit is equivalent to the integral $\int_{0}^{1} \frac{1}{1+x} dx$. Note that $\int_{1}^{2} \frac{1}{x} dx$ has the same value, $\ln(2)$, but it corresponds to the sum $\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{1+i/n} \cdot \frac{1}{n}$, which is not the original expression.
18. Suppose $f$ is an integrable function and $\lim_{n\to\infty} \sum_{i=1}^n f(c_i) \Delta x_i = I$ for any choice of points $c_i$ in the $i$-th subinterval. If the interval of integration is $[2, 6]$ and a midpoint Riemann sum with 4 equal subintervals gives a value of 10, what is $\Delta x$ for this sum?
    Answer: 1 This question tests the understanding of the components of a Riemann sum, specifically $\Delta x$. The other information provided (the value of the sum is 10, it's a midpoint sum, the limit exists) is extraneous to the question being asked. The interval is $[2, 6]$ and there are $n=4$ equal subintervals. The width of each subinterval is calculated as $\Delta x = \frac{b-a}{n} = \frac{6-2}{4} = \frac{4}{4} = 1$.
19. Which of the following expressions represents a left Riemann sum approximation for $\int_3^7 \ln(x) \,dx$ with $n=8$ subintervals?
    Answer: $0.5 \sum_{i=0}^{7} \ln(3 + 0.5i)$ For the integral $\int_3^7 \ln(x) \,dx$ with $n=8$, the width of each subinterval is $\Delta x = \frac{7-3}{8} = \frac{4}{8} = 0.5$. The partition points are $x_0=3, x_1=3.5, x_2=4, \dots, x_8=7$. A left Riemann sum uses the left endpoints of the subintervals: $x_0, x_1, \dots, x_7$. The formula for the left endpoints is $x_i = a + i\Delta x = 3 + 0.5i$. The sum is $\sum_{i=0}^{n-1} f(x_i) \Delta x = \sum_{i=0}^{7} \ln(x_i) (0.5) = 0.5 \sum_{i=0}^{7} \ln(3 + 0.5i)$. B) represents the right Riemann sum. C) represents the midpoint Riemann sum. D) incorrectly uses the total interval width (4) as the multiplier and also as part of the function argument, confusing the setup.
20. The limit $\lim_{n \to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2}$ can be expressed as a definite integral. Which integral is it?
    Answer: $\int_0^1 \frac{1}{1+x^2} \,dx$ To convert the sum into the form $\lim \sum f(x_i) \Delta x$, we need to factor out $\frac{1}{n}$ (for $\Delta x$). We can rewrite the term inside the summation by dividing the numerator and denominator by $n^2$: $\frac{n}{n^2+k^2} = \frac{n/n^2}{(n^2+k^2)/n^2} = \frac{1/n}{1+(k/n)^2}$. The summation becomes $\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{1+(k/n)^2} \cdot \frac{1}{n}$. Now we can identify the components for an integral on $[0, 1]$: $\Delta x = 1/n$ and $x_k = k/n$. The function is $f(x) = \frac{1}{1+x^2}$. The integral is $\int_0^1 \frac{1}{1+x^2} \,dx$.

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