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Integral Calculus: JEE Main Mathematics Question with Solution

Q: Why do we split the integral at x = 1?

Because when and when . The absolute value definition requires when .

\int\limits_{\frac{1}{3}}^3 |\log_{\text{e}} \text{x}|\text{dx} = \frac{\text{m}}{\text{n}}\log_{\text{e}}\left(\frac{\text{n}^2}{\text{e}}\right)

, where

\text{m}

and

\text{n}

are coprime natural numbers, then

\text{m}^2 + \text{n}^2 - 5

is equal to

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Step-by-step solutionView

Correct answer

The value of $m^2 + n^2 - 5$ is $20$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Integral $I = \int_{1/3}^3 |\ln x|\,dx = \frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ with $m, n \in \mathbb{N}$ and $\gcd(m, n) = 1$ .

02goal

Evaluate the integral, match it to $\frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ to determine $m$ and $n$ , and compute $m^2 + n^2 - 5$ .

03approach

Split the domain at $x = 1$ , where $\ln x$ changes sign. For $x \in [1/3, 1]$ , $|\ln x| = -\ln x$ , and for $x \in [1, 3]$ , $|\ln x| = \ln x$ . Then use $\int \ln x\,dx = x\ln x - x$ .

04execute

Split the integral: $I = -\int_{1/3}^1 \ln x\,dx + \int_1^3 \ln x\,dx$ Antiderivative of $\ln x$ is $F(x) = x\ln x - x$ . Evaluating: $-[x\ln x - x]_{1/3}^1 = -[(0 - 1) - (\frac{1}{3}\ln(1/3) - \frac{1}{3})] = -[-1 + \frac{1}{3}\ln 3 + \frac{1}{3}] = -[-\frac{2}{3} + \frac{1}{3}\ln 3] = \frac{2}{3} - \frac{1}{3}\ln 3$ $[x\ln x - x]_1^3 = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$ Adding both parts: $I = \left(\frac{2}{3} - \frac{1}{3}\ln 3\right) + (3\ln 3 - 2) = \frac{8}{3}\ln 3 - \frac{4}{3} = \frac{4}{3}(2\ln 3 - 1) = \frac{4}{3}\ln\left(\frac{3^2}{e}\right)$

05execute

Equating $\frac{4}{3}\ln\left(\frac{3^2}{e}\right)$ with $\frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ gives $m = 4$ and $n = 3$ , which are coprime natural numbers. Compute the required expression: $m^2 + n^2 - 5 = 4^2 + 3^2 - 5 = 16 + 9 - 5 = 20$

✓verify

Check values: $m=4, n=3 \implies \gcd(4,3)=1$ . The formula inside the logarithm has $n^2/e = 9/e$ , exactly matching $\ln(9) - 1 = 2\ln 3 - 1$ .

Hints that build this answer step by step

How should the integral $\int_{1/3}^3 |\ln x|\,dx$ be split to remove the absolute value?
$-\int_{1/3}^1 \ln x\,dx + \int_1^3 \ln x\,dx$
Using the antiderivative $\int \ln x\,dx = x\ln x - x$ , what is the evaluated value of the integral?
$\frac{8}{3}\ln 3 - \frac{4}{3}$
Rewriting $\frac{8}{3}\ln 3 - \frac{4}{3}$ in the form $\frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ , what are the coprime values of $m$ and $n$ ?
$m = 4$ and $n = 3$
With $m = 4$ and $n = 3$ , what is the final value of $m^2 + n^2 - 5$ ?
$20$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we split the integral at x = 1?

Because $\ln x < 0$ when $0 < x < 1$ and $\ln x \ge 0$ when $x \ge 1$ . The absolute value definition requires $|u| = -u$ when $u < 0$ .

Answer

The value of $m^2 + n^2 - 5$ is $20$ .

Why each option works or fails

Step-by-step solution

given: Integral $I = \int_{1/3}^3 |\ln x|\,dx = \frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ with $m, n \in \mathbb{N}$ and $\gcd(m, n) = 1$ .
goal: Evaluate the integral, match it to $\frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ to determine $m$ and $n$ , and compute $m^2 + n^2 - 5$ .
approach: Split the domain at $x = 1$ , where $\ln x$ changes sign. For $x \in [1/3, 1]$ , $|\ln x| = -\ln x$ , and for $x \in [1, 3]$ , $|\ln x| = \ln x$ . Then use $\int \ln x\,dx = x\ln x - x$ .
execute: Split the integral: $I = -\int_{1/3}^1 \ln x\,dx + \int_1^3 \ln x\,dx$ Antiderivative of $\ln x$ is $F(x) = x\ln x - x$ . Evaluating: $-[x\ln x - x]_{1/3}^1 = -[(0 - 1) - (\frac{1}{3}\ln(1/3) - \frac{1}{3})] = -[-1 + \frac{1}{3}\ln 3 + \frac{1}{3}] = -[-\frac{2}{3} + \frac{1}{3}\ln 3] = \frac{2}{3} - \frac{1}{3}\ln 3$ $[x\ln x - x]_1^3 = (3\ln 3 - 3) - (0 - 1) = 3\ln 3 - 2$ Adding both parts: $I = \left(\frac{2}{3} - \frac{1}{3}\ln 3\right) + (3\ln 3 - 2) = \frac{8}{3}\ln 3 - \frac{4}{3} = \frac{4}{3}(2\ln 3 - 1) = \frac{4}{3}\ln\left(\frac{3^2}{e}\right)$
execute: Equating $\frac{4}{3}\ln\left(\frac{3^2}{e}\right)$ with $\frac{m}{n}\ln\left(\frac{n^2}{e}\right)$ gives $m = 4$ and $n = 3$ , which are coprime natural numbers. Compute the required expression: $m^2 + n^2 - 5 = 4^2 + 3^2 - 5 = 16 + 9 - 5 = 20$
verify: Check values: $m=4, n=3 \implies \gcd(4,3)=1$ . The formula inside the logarithm has $n^2/e = 9/e$ , exactly matching $\ln(9) - 1 = 2\ln 3 - 1$ .

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

The value of $m^2 + n^2 - 5$ is $20$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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