Calculus 2 Quiz: Riemann Sums And Notation

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Riemann Sums And NotationQuestion 1 of 20

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 $[1, 5]$ ?

The right Riemann sum underestimates the integral by exactly 3 units

The right Riemann sum overestimates the integral by some unknown amount

The right Riemann sum underestimates the integral by some unknown amount

The right Riemann sum equals the exact integral value of 12

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Calculus 2 Quiz

Calculus 2 Quiz: Riemann Sums And Notation

Practice Riemann Sums And Notation in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Riemann Sums And Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The right Riemann sum underestimates the integral by exactly 3 units
The right Riemann sum overestimates the integral by some unknown amount
The right Riemann sum underestimates the integral by some unknown amount (correct answer)
The right Riemann sum equals the exact integral value of 12

Explanation: 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.

Question 2

Let $S_n = \sum_{i=1}^n (\frac{2i}{n^2} - \frac{3}{n})$ . What is the value of $\lim_{n\to\infty} S_n$ ?

-3
-2 (correct answer)
1
0

Explanation: 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

Question 3

Consider the function $f(x) = x^2 + 1$ on the interval $[0, 4]$ . If we use a left Riemann sum with $n = 4$ subintervals, which expression correctly represents the sum in summation notation?

$\sum_{i=0}^{3} f(i) \cdot 1$ (correct answer)
$\sum_{i=1}^{4} f(i-1) \cdot 1$
$\sum_{i=0}^{4} f(i) \cdot 1$
$\sum_{i=1}^{3} f(i) \cdot 1$

Explanation: For a left Riemann sum on [0,4] with n=4 subintervals, Δx = (4-0)/4 = 1. The left endpoints are x₀=0, x₁=1, x₂=2, x₃=3. So we evaluate f at these points and multiply by Δx=1. This gives us ∑(i=0 to 3) f(i)·1. Choice B is equivalent but unnecessarily complicated. Choice C includes an extra term f(4), and choice D misses the first and last terms.

Question 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?

$L_n < T_n < \int_a^b f(x) dx < M_n < R_n$
$R_n < \int_a^b f(x) dx < M_n < T_n < L_n$
$R_n < M_n < \int_a^b f(x) dx < T_n < L_n$ (correct answer)
$M_n < R_n < \int_a^b f(x) dx < L_n < T_n$

Explanation: 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

Question 5

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?

7/6
13/12
11/2
11/6 (correct answer)

Explanation: 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.

Question 6

Consider the definite integral $I = \int_1^3 \frac{1}{x} dx$ . Let $R_n$ be the right-hand Riemann sum approximation and $L_n$ be the left-hand Riemann sum approximation. Which statement correctly describes the relationship between $R_{10}$ , $L_{10}$ , and $I$ ?

$L_{10} < R_{10} < I$
$R_{10} < I < L_{10}$ (correct answer)
$R_{10} < L_{10} < I$
$I < R_{10} < L_{10}$

Explanation: The function

f(x) = 1/x

is a strictly decreasing function on the interval

[1, 3]

. For any strictly decreasing function, the left-hand Riemann sum

L_n

will always be an overestimate of the true integral, because each rectangle's height is determined by the highest point in its subinterval. Conversely, the right-hand Riemann sum

R_n

will always be an underestimate, as each rectangle's height is determined by the lowest point. Therefore, for any

n

, including

n=10

, we must have

R_{10} < I < L_{10}

Question 7

A Riemann sum for a function $f(x)$ on the interval $[2, 10]$ is given by $S_n = \sum_{i=1}^n \left(2 + \frac{8(i-1)}{n}\right)^2 \frac{8}{n}$ . As $n \to \infty$ , what is the value of the corresponding definite integral?

336
328
992/3 (correct answer)
1000/3

Explanation: The Riemann sum is

S_n = \sum_{i=1}^n f(x_i^*) \Delta x

. From the term

\frac{8}{n}

, we identify

\Delta x = \frac{10-2}{n} = \frac{8}{n}

. The sample points are

x_i^* = 2 + \frac{8(i-1)}{n}

. This is of the form

a + (i-1)\Delta x

, which represents the left endpoint of the i-th interval. The function being evaluated is

f(x) = x^2

. Therefore, the sum represents the definite integral

\int_2^{10} x^2 dx

. Evaluating this gives

[\frac{x^3}{3}]_2^{10} = \frac{10^3}{3} - \frac{2^3}{3} = \frac{1000 - 8}{3} = \frac{992}{3}

Question 8

Which of the following summation expressions, when the limit is taken as $n \to \infty$ , represents $\int_0^1 e^{-2x} dx$ ?

$\sum_{i=1}^n e^{-2i/n} \frac{2}{n}$
$\sum_{i=1}^n e^{-i/n} \frac{1}{n}$
$\sum_{i=1}^n e^{-2i/n} \frac{1}{n}$ (correct answer)
$\sum_{i=0}^{n-1} e^{-2(i+1)/n} \frac{2}{n}$

Explanation: 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

Question 9

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?

32
8
4
16 (correct answer)

Explanation: 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

[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

Question 10

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?

A square with side length 1
A right triangle with legs of length 1
A semi-circle with radius 1
A quarter-circle with radius 1 (correct answer)

Explanation: 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

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.

Question 11

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, $R_n$ ?

Change the index of the sample point from $x_{i-1}$ to $x_{i+1}$ .
Change the summation to run from $i=0$ to $n$ .
Change $\Delta x$ to $(b-a)/2n$ .
Change the index of the sample point from $x_{i-1}$ to $x_i$ . (correct answer)

Explanation: 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.

Question 12

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?

Interval $[1, 3]$ using left endpoints with 4 equal subintervals
Interval $[1, 3]$ using right endpoints with 4 equal subintervals (correct answer)
Interval $[1.5, 3.5]$ using left endpoints with 4 equal subintervals
Interval $[1, 3]$ using midpoints with 4 equal subintervals

Explanation: 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.

Question 13

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?

$\int_0^2 \sqrt{1 + x^2} \, dx$ (correct answer)
$\int_1^3 \sqrt{1 + x^2} \, dx$
$\int_0^2 \sqrt{1 + 4x^2} \, dx$
$\int_1^2 \sqrt{1 + x^2} \, dx$

Explanation: 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.

Question 14

A function $p(x)$ is continuous and positive on $[0, 3]$ . The midpoint Riemann sum with $n = 6$ subintervals gives an approximation of 15.7 for $\int_0^3 p(x) \, dx$ . If we know that $p(x)$ is concave up on $[0, 3]$ , what can we conclude about the exact value of the integral?

The exact value is greater than 15.7 because midpoint sums overestimate for concave up functions
The exact value is less than 15.7 because midpoint sums underestimate for concave up functions (correct answer)
The exact value equals 15.7 because midpoint sums are exact for all continuous functions
We cannot determine the relationship without knowing if the function is increasing or decreasing

Explanation: For a function that is concave up, the midpoint Riemann sum underestimates the true value of the definite integral. This is because when a function is concave up, the function lies above its secant lines, and the midpoint rule approximates the area using rectangles whose heights correspond to points on the curve that lie below the average value of the function on each subinterval. Choice A has the wrong direction. Choice C is incorrect since midpoint sums are approximations. Choice D is wrong because concavity alone determines the relationship for midpoint sums.

Question 15

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?

The interval should be $[-1, 3]$ based on the given sum notation
The sum should use right endpoints, not left endpoints for this interval
The sum correctly represents a left Riemann sum for $\int_{-1}^{2} f(x) \, dx$ (correct answer)
The width $\frac{3}{n}$ is incorrect for the given interval length

Explanation: 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.

Question 16

What is the exact value of $\lim_{n\to\infty} \sum_{i=1}^n \frac{2}{n}\left(1 + \frac{i}{n}\right)$ ?

3 (correct answer)
2
4
1

Explanation: 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

[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

Question 17

The sum $\sum_{i=1}^n (i^3 - (i-1)^3)$ is an example of a telescoping series. What is its value?

$n^3 - 1$
$(n+1)^3 - 1$
$n^3$ (correct answer)
$1$

Explanation: This is a telescoping sum. Let's write out the first few terms and the last term. For

i=1

1^3 - 0^3

. For

i=2

2^3 - 1^3

. For

i=3

3^3 - 2^3

. ... For

i=n

n^3 - (n-1)^3

. When we add these terms, the second part of each term cancels with the first part of the next term:

(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + ... + (n^3 - (n-1)^3)

. The

-1^3

cancels with

1^3

-2^3

cancels with

2^3

, and so on, until

-(n-1)^3

cancels. The only terms that do not cancel are

-0^3

from the first term and

n^3

from the last term. The sum is

n^3 - 0^3 = n^3

. This identity is useful for deriving the formula for

\sum i^2

Question 18

The definite integral $\int_2^5 (x^2 - 1) dx$ is expressed as the limit of a Riemann sum. Which of the following limits correctly represents this integral using right endpoints?

$\lim_{n\to\infty} \sum_{i=1}^n \left( \left(2 + \frac{3i}{n}\right)^2 - 1 \right) \frac{3}{n}$ (correct answer)
$\lim_{n\to\infty} \sum_{i=1}^n \left( \left(\frac{3i}{n}\right)^2 - 1 \right) \frac{3}{n}$
$\lim_{n\to\infty} \sum_{i=1}^n \left( \left(2 + \frac{5i}{n}\right)^2 - 1 \right) \frac{5}{n}$
$\lim_{n\to\infty} \sum_{i=1}^n \left( \left(2 + \frac{3(i-1)}{n}\right)^2 - 1 \right) \frac{3}{n}$

Explanation: For the integral

\int_a^b f(x) dx

, the limit of the right-hand Riemann sum is

\lim_{n\to\infty} \sum_{i=1}^n f(x_i) \Delta x

. Here,

a=2

b=5

, and

f(x) = x^2 - 1

. The width of each subinterval is

\Delta x = \frac{b-a}{n} = \frac{5-2}{n} = \frac{3}{n}

. The right endpoint of the i-th subinterval is

x_i = a + i\Delta x = 2 + i\frac{3}{n}

. Substituting these into the formula gives

f(x_i) = (2 + \frac{3i}{n})^2 - 1

. Therefore, the correct limit expression is

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

Question 19

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?

8 (correct answer)
0
-8
7

Explanation: 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.

Question 20

The limit $\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i^2}{n^3}$ can be expressed as a definite integral. What is the value of this integral?

1/3 (correct answer)
1/2
1
1/4

Explanation: First, we rewrite the expression to fit the form of a Riemann sum,

\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x

. We can write

\frac{i^2}{n^3} = \left(\frac{i}{n}\right)^2 \frac{1}{n}

. This matches the form with

\Delta x = 1/n

and sample points

x_i = i/n

for a right-hand sum on the interval

[0, 1]

. The function is

f(x) = x^2

. Therefore, the limit is equivalent to the definite integral

\int_0^1 x^2 dx

. Evaluating this integral gives

\left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}