Why does correspond to ?

. Hence, the index runs from to .

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Integral Calculus: JEE Main Mathematics Question with Solution

\lim_{\mathrm{n}\to\infty} \frac{3}{\mathrm{n}} \left\{4 + \left(2+\frac{1}{\mathrm{n}}\right)^2 + \left(2+\frac{2}{\mathrm{n}}\right)^2 + \dots + \left(3-\frac{1}{\mathrm{n}}\right)^2\right\}

is equal to

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12

\frac{19}{3}

0

19

Step-by-step solutionView

Correct answer

The limit evaluates to $19$ by rewriting the sum as a definite integral $3 \int_0^1 (2+x)^2 \, dx$ .

Option analysis

Why each option works or fails

A ·

12

Evaluating the integral by ignoring the squared term or miscalculating $\int_2^3 x^2 dx$ as something yielding $4$ . Compute $\int_2^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_2^3 = \frac{27 - 8}{3} = \frac{19}{3}$ , then multiply by the outer factor of $3$ to get $19$ .

B ·

\frac{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

None. This is the correct answer. Correctly expressed the limit as $3 \int_0^1 (2+x)^2 \, dx = 3 \left[\frac{(2+x)^3}{3}\right]_0^1 = 3^3 - 2^3 = 27 - 8 = 19$ .

Reviewed route

Solution

StepWorking

01given

Given expression: $S = \lim_{n\to\infty} \frac{3}{n} \left\{ 4 + \left(2+\frac{1}{n}\right)^2 + \left(2+\frac{2}{n}\right)^2 + \dots + \left(3-\frac{1}{n}\right)^2 \right\}$ . 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 $\int_0^1 f(x)\,dx$ where $x = r/n$ and $dx = 1/n$ .

04execute

Convert to definite integral: $S = \int_0^1 3(2+x)^2\,dx = \left[ (2+x)^3 \right]_0^1 = (2+1)^3 - (2+0)^3 = 27 - 8 = 19$ .

✓verify

Alternatively, substitute $u = 2+x$ , $du = dx$ , with limits from $u=2$ to $u=3$ : $\int_2^3 3u^2\,du = [u^3]_2^3 = 27 - 8 = 19$ . Both approaches match.

Hints that build this answer step by step

How can the given expression inside the limit be written in sigma notation?
$3 \lim_{n\to\infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2$
Which definite integral represents $\lim_{n\to\infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2$ ?
$\int_0^1 (2+x)^2 \, dx$
What is the final value of $3 \int_0^1 (2+x)^2 \, dx$ ?
$19$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $3 \int_0^1 (2+x)^2 \, dx$ .

Why each option works or fails

A: $12$ - Evaluating the integral by ignoring the squared term or miscalculating $\int_2^3 x^2 dx$ as something yielding $4$ . Compute $\int_2^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_2^3 = \frac{27 - 8}{3} = \frac{19}{3}$ , then multiply by the outer factor of $3$ to get $19$ .
B: $\frac{19}{3}$ - 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: $19$ - None. This is the correct answer. Correctly expressed the limit as $3 \int_0^1 (2+x)^2 \, dx = 3 \left[\frac{(2+x)^3}{3}\right]_0^1 = 3^3 - 2^3 = 27 - 8 = 19$ .

Step-by-step solution

given: Given expression: $S = \lim_{n\to\infty} \frac{3}{n} \left\{ 4 + \left(2+\frac{1}{n}\right)^2 + \left(2+\frac{2}{n}\right)^2 + \dots + \left(3-\frac{1}{n}\right)^2 \right\}$ . Note that $4 = (2+0/n)^2$ and $(3-1/n) = (2 + (n-1)/n)$ .
goal: Evaluate the limit as $\frac{19}{3}$ 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 $\int_0^1 f(x)\,dx$ where $x = r/n$ and $dx = 1/n$ .
execute: Convert to definite integral: $S = \int_0^1 3(2+x)^2\,dx = \left[ (2+x)^3 \right]_0^1 = (2+1)^3 - (2+0)^3 = 27 - 8 = 19$ .
verify: Alternatively, substitute $u = 2+x$ , $du = dx$ , with limits from $u=2$ to $u=3$ : $\int_2^3 3u^2\,du = [u^3]_2^3 = 27 - 8 = 19$ . Both approaches match.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The limit evaluates to $19$ by rewriting the sum as a definite integral $3 \int_0^1 (2+x)^2 \, dx$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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