JEE MainMathematics

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Definite Integral as the Limit of a Riemann Sum

Q: Why does (3 - 1/n)^2 correspond to r = n-1?

3 - 1/n = 2 + 1 - 1/n = 2 + (n-1)/n. Hence, the index r runs from 0 to n-1.

lim as n→∞ of 3/n {4 + (2+1/n)^2 + (2+2/n)^2 + … + (3-1/n)^2} is equal to

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A12B19/3C0D19

Step-by-step solutionView

Correct answer

The limit evaluates to $19$ by rewriting the sum as a definite integral $3 \int_{0}^{1} (2 + x)^{2} d x$ .

Option analysis

Why each option works or fails

A · 12

Evaluating the integral by ignoring the squared term or miscalculating $\int_{2}^{3} x^{2} d x$ as something yielding $4$ . Compute $\int_{2}^{3} x^{2} d x = {[\frac{x^{3}}{3}]}_{2}^{3} = \frac{27 - 8}{3} = \frac{19}{3}$ , then multiply by the outer factor of $3$ to get $19$ .

B · 19/3

Forgetting the factor of $3$ multiplying the summation expression. Retain the constant multiplier $3$ from $\frac{3}{n}$ , multiplying the resulting integral $\frac{19}{3}$ by $3$ to obtain $19$ .

C · 0

Assuming that the factor of $\frac{3}{n}$ forces the entire limit to $0$ without recognizing the sum contains $n$ terms. Recognize that a sum of $n$ terms scaled by $\frac{1}{n}$ forms a Riemann sum converging to a non-zero definite integral.

D · 19

This is the correct answer. Correctly expressed the limit as $3 \int_{0}^{1} (2 + x)^{2} d x = 3 {[\frac{(2 + x)^{3}}{3}]}_{0}^{1} = 3^{3} - 2^{3} = 27 - 8 = 19$ .

Reviewed route

Solution

StepWorking

01Given

Given expression: $S = \lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ . Note that $4 = (2 + 0 / n)^{2}$ and $(3 - 1 / n) = (2 + (n - 1) / n)$ .

02Goal

Evaluate the limit as $n \to \infty$ to find the numerical value.

03Approach

Express the series as a Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} f (r / n)$ which converts into the definite integral $\int_{0}^{1} f (x) d x$ where $x = r / n$ and $d x = 1 / n$ .

04Execute

Convert to definite integral: $S = \int_{0}^{1} 3 (2 + x)^{2} d x = {[(2 + x)^{3}]}_{0}^{1} = (2 + 1)^{3} - (2 + 0)^{3} = 27 - 8 = 19$ .

✓Verify

Alternatively, substitute $u = 2 + x$ , $d u = d x$ , with limits from $u = 2$ to $u = 3$ : $\int_{2}^{3} 3 u^{2} d u = [u^{3}]_{2}^{3} = 27 - 8 = 19$ . Both approaches match.

Hints that build this answer step by step

How can the given expression inside the limit be written in sigma notation?
$3 \lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} {(2 + \frac{r}{n})}^{2}$
Which definite integral represents $\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} {(2 + \frac{r}{n})}^{2}$ ?
$\int_{0}^{1} (2 + x)^{2} d x$
What is the final value of $3 \int_{0}^{1} (2 + x)^{2} d x$ ?
$19$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why does

(3 - 1 / n)^{2}

correspond to

r = n - 1

$3 - 1 / n = 2 + 1 - 1 / n = 2 + (n - 1) / n$ . Hence, the index $r$ runs from $0$ to $n - 1$ .

Answer

The limit evaluates to $19$ by rewriting the sum as a definite integral $3 \int_{0}^{1} (2 + x)^{2} d x$ .

Why each option works or fails

A: 12 - Evaluating the integral by ignoring the squared term or miscalculating $\int_{2}^{3} x^{2} d x$ as something yielding $4$ . Compute $\int_{2}^{3} x^{2} d x = {[\frac{x^{3}}{3}]}_{2}^{3} = \frac{27 - 8}{3} = \frac{19}{3}$ , then multiply by the outer factor of $3$ to get $19$ .
B: 19/3 - Forgetting the factor of $3$ multiplying the summation expression. Retain the constant multiplier $3$ from $\frac{3}{n}$ , multiplying the resulting integral $\frac{19}{3}$ by $3$ to obtain $19$ .
C: 0 - Assuming that the factor of $\frac{3}{n}$ forces the entire limit to $0$ without recognizing the sum contains $n$ terms. Recognize that a sum of $n$ terms scaled by $\frac{1}{n}$ forms a Riemann sum converging to a non-zero definite integral.
D · correct: 19 - This is the correct answer. Correctly expressed the limit as $3 \int_{0}^{1} (2 + x)^{2} d x = 3 {[\frac{(2 + x)^{3}}{3}]}_{0}^{1} = 3^{3} - 2^{3} = 27 - 8 = 19$ .

Step-by-step solution

Given: Given expression: $S = \lim_{n \to \infty} \frac{3}{n} {4 + {(2 + \frac{1}{n})}^{2} + {(2 + \frac{2}{n})}^{2} + \dots + {(3 - \frac{1}{n})}^{2}}$ . Note that $\frac{1}{n}$ and $(3 - 1 / n) = (2 + (n - 1) / n)$ .
Goal: Evaluate the limit as $n \to \infty$ to find the numerical value.
Approach: Express the series as a Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} f (r / n)$ which converts into the definite integral $\int_{0}^{1} f (x) d x$ where $x = r / n$ and $d x = 1 / n$ .
Execute: Convert to definite integral: $S = \int_{0}^{1} 3 (2 + x)^{2} d x = {[(2 + x)^{3}]}_{0}^{1} = (2 + 1)^{3} - (2 + 0)^{3} = 27 - 8 = 19$ .
Verify: Alternatively, substitute $u = 2 + x$ , $d u = d x$ , with limits from $u = 2$ to $\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n - 1} f (r / n)$ : $\int_{2}^{3} 3 u^{2} d u = [u^{3}]_{2}^{3} = 27 - 8 = 19$ . Both approaches match.

Definite Integral as the Limit of a Riemann Sum

What feels right?

The limit evaluates to $19$ by rewriting the sum as a definite integral $3 \int_{0}^{1} (2 + x)^{2} d x$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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