JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why do the integration limits become 0 to 1?

Because the lower limit is , and the upper limit is .

\lim_{n \to \infty} \left[\frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \dots + \frac{1}{2n}\right]

is equal to

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\log_e 2

\log_e\left(\frac{3}{2}\right)

\log_e\left(\frac{2}{3}\right)

0

Step-by-step solutionView

Correct answer

The limit equals $\int_0^1 \frac{1}{1+x}\,dx = \log_e 2$ .

Option analysis

Why each option works or fails

A ·

\log_e 2

None. The student correctly converts the limit of the Riemann sum into a definite integral $\int_0^1 \frac{dx}{1+x} = [\ln(1+x)]_0^1 = \ln 2$ . Correctly expressed the sum in terms of $\frac{r}{n}$ , factored out $\frac{1}{n}$ , and evaluated the resulting definite integral.

B ·

\log_e\left(\frac{3}{2}\right)

The student mistakenly sets the upper integration limit or substitution range incorrectly, evaluating $\ln(1+x)$ from $x=1/2$ to $1$ or similar, leading to $\ln(3/2)$ . Identify the running index $r$ from $1$ to $n$ , which gives limits of integration from $\lim \frac{1}{n} = 0$ to $\lim \frac{n}{n} = 1$ .

C ·

\log_e\left(\frac{2}{3}\right)

The student inverts the argument of the logarithm, possibly evaluating $[\ln(1+x)]$ as $\ln(1) - \ln(1+1) = -\ln 2$ or miscomputing $\ln(1+1/2) - \ln(2) = \ln(2/3)$ . Evaluate the antiderivative properly at the upper limit minus the lower limit: $[\ln(1+x)]_0^1 = \ln(2) - \ln(1) = \ln 2$ .

D ·

0

The student takes the limit of each term individually as $n \to \infty$ , getting $\lim_{n \to \infty} \frac{1}{r+n} = 0$ , and concludes the sum of $n$ such terms is 0. Because the number of terms grows with $n$ , you cannot distribute the limit across the sum; formulate it as a Riemann sum instead.

Reviewed route

Solution

StepWorking

01given

The given expression is $S = \lim_{n \to \infty} \left[\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{n+n}\right]$ .

02goal

Evaluate the limit as $n \to \infty$ using Riemann sum integration.

03approach

Factor out $\frac{1}{n}$ to express the general term as a function of $\frac{r}{n}$ , and convert the Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum f\left(\frac{r}{n}\right)$ into $\int_0^1 f(x)\,dx$ .

04execute

$S = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n+r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1 + \frac{r}{n}} = \int_0^1 \frac{1}{1+x}\,dx = [\ln(1+x)]_0^1 = \ln 2 - \ln 1 = \ln 2$ .

✓verify

For finite $n$ , $\sum_{r=1}^n \frac{1}{n+r} = H_{2n} - H_n \approx (\ln(2n) + \gamma) - (\ln n + \gamma) = \ln 2$ . The result matches.

✓ 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: 7 September 2026

Quick checks

Students also ask

Why do the integration limits become 0 to 1?

Because the lower limit is $\lim_{n \to \infty} \frac{r_{\min}}{n} = \lim_{n \to \infty} \frac{1}{n} = 0$ , and the upper limit is $\lim_{n \to \infty} \frac{r_{\max}}{n} = \lim_{n \to \infty} \frac{n}{n} = 1$ .

Answer

The limit equals $\int_0^1 \frac{1}{1+x}\,dx = \log_e 2$ .

Why each option works or fails

A · correct: $\log_e 2$ - None. The student correctly converts the limit of the Riemann sum into a definite integral $\int_0^1 \frac{dx}{1+x} = [\ln(1+x)]_0^1 = \ln 2$ . Correctly expressed the sum in terms of $\frac{r}{n}$ , factored out $\frac{1}{n}$ , and evaluated the resulting definite integral.
B: $\log_e\left(\frac{3}{2}\right)$ - The student mistakenly sets the upper integration limit or substitution range incorrectly, evaluating $\ln(1+x)$ from $x=1/2$ to $1$ or similar, leading to $\ln(3/2)$ . Identify the running index $r$ from $1$ to $n$ , which gives limits of integration from $\lim \frac{1}{n} = 0$ to $\lim \frac{n}{n} = 1$ .
C: $\log_e\left(\frac{2}{3}\right)$ - The student inverts the argument of the logarithm, possibly evaluating $[\ln(1+x)]$ as $\ln(1) - \ln(1+1) = -\ln 2$ or miscomputing $\ln(1+1/2) - \ln(2) = \ln(2/3)$ . Evaluate the antiderivative properly at the upper limit minus the lower limit: $[\ln(1+x)]_0^1 = \ln(2) - \ln(1) = \ln 2$ .
D: $0$ - The student takes the limit of each term individually as $n \to \infty$ , getting $\lim_{n \to \infty} \frac{1}{r+n} = 0$ , and concludes the sum of $n$ such terms is 0. Because the number of terms grows with $n$ , you cannot distribute the limit across the sum; formulate it as a Riemann sum instead.

Step-by-step solution

given: The given expression is $S = \lim_{n \to \infty} \left[\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{n+n}\right]$ .
goal: Evaluate the limit as $n \to \infty$ using Riemann sum integration.
approach: Factor out $\frac{1}{n}$ to express the general term as a function of $\frac{r}{n}$ , and convert the Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum f\left(\frac{r}{n}\right)$ into $\int_0^1 f(x)\,dx$ .
execute: $S = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n+r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1 + \frac{r}{n}} = \int_0^1 \frac{1}{1+x}\,dx = [\ln(1+x)]_0^1 = \ln 2 - \ln 1 = \ln 2$ .
verify: For finite $n$ , $\sum_{r=1}^n \frac{1}{n+r} = H_{2n} - H_n \approx (\ln(2n) + \gamma) - (\ln n + \gamma) = \ln 2$ . The result matches.

Integral Calculus: JEE Main Mathematics Question with Solution

The limit equals ∫0111+x dx=log⁡e2\int_0^1 \frac{1}{1+x}\,dx = \log_e 2∫01​1+x1​dx=loge​2.

01given

02goal

03approach

04execute

✓verify

Students also ask

The limit equals $\int_0^1 \frac{1}{1+x}\,dx = \log_e 2$ .