JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite integral of an even function involving absolute values

The value of the integral integral from -2 to 2 of (| x^3 + x |)/((e^(|x||x|) + 1)) dx is equal to:

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A5e^2B3e^(-2)C4D6

Step-by-step solutionView

Correct answer

Because the integrand is an even function, the integral over $[- 2, 2]$ simplifies to $2 \int_{0}^{2} \frac{x^{3} + x}{e^{x^{2}} + 1} d x$ , which evaluates to $6$ after applying the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ or direct substitution.

Option analysis

Why each option works or fails

A · 5e^2

Believing that integrating the exponential in the denominator leaves an unevaluated term proportional to $e^{2}$ . Notice that multiplying or pairing terms using King's property or substitution cancels out the exponential term entirely, leaving a simple polynomial.

B · 3e^(-2)

Assuming that the denominator $e^{x^{2}} + 1$ introduces negative exponential factors into the final value. Substitute $u = x^{2}$ to get $d u = 2 x d x$ , which transforms the integrand into $\frac{u + 1}{e^{u} + 1}$ , where the exponential disappears after using reflection symmetry or paired integrals.

C · 4

Integrating without multiplying by $2$ when reducing the integral of the even function from $[- 2, 2]$ to $[0, 2]$ , or incorrectly evaluating ${[\frac{u^{2}}{2} + u]}_{0}^{4}$ . Ensure the factor of $2$ from the symmetric interval property $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ is properly included throughout the calculation.

D · 6

The integral is evaluated correctly by exploiting symmetry. The integrand is even, so the integral equals $2 \int_{0}^{2} \frac{x (x^{2} + 1)}{e^{x^{2}} + 1} d x$ . Letting $t = x^{2}$ gives $\int_{0}^{4} \frac{t + 1}{e^{t} + 1} d t$ . Using the reflection property $\int_{0}^{a} f (t) d t = \int_{0}^{a} f (a - t) d t$ cancels the denominator and yields $6$ .

Reviewed route

Solution

StepWorking

01Given

The integral is $I = \int_{- 2}^{2} \frac{∣ x^{3} + x ∣}{e^{x ∣ x ∣} + 1} d x$ (noting the standard typo in the denominator of the stem where $e^{∣ x ∣ ∣ x ∣}$ is actually $e^{x ∣ x ∣}$ , consistent with standard JEE papers).

02Goal

Evaluate $I = \int_{- 2}^{2} f (x) d x$ where $f (x) = \frac{∣ x^{3} + x ∣}{e^{x ∣ x ∣} + 1}$ .

03Approach

Use King's property for symmetric limits $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x$ . Notice that the numerator $∣ x^{3} + x ∣$ is an even function, while the exponent $x ∣ x ∣$ in the denominator is an odd function.

04Execute

For $x \in [0, 2]$ , we have $∣ x^{3} + x ∣ = x^{3} + x$ and $x ∣ x ∣ = x^{2}$ . Hence, $f (x) = \frac{x^{3} + x}{e^{x^{2}} + 1}$ . For $f (- x)$ , the numerator is $∣ (- x)^{3} + (- x) ∣ = ∣ - (x^{3} + x) ∣ = x^{3} + x$ , and the exponent is $(- x) ∣ - x ∣ = - x ∣ x ∣ = - x^{2}$ . Thus, $f (- x) = \frac{x^{3} + x}{e^{- x^{2}} + 1} = \frac{(x^{3} + x) e^{x^{2}}}{1 + e^{x^{2}}}$ .

05Execute

Add $f (x)$ and $f (- x)$ on $[0, 2]$ : $f (x) + f (- x) = (x^{3} + x) [\frac{1}{e^{x^{2}} + 1} + \frac{e^{x^{2}}}{e^{x^{2}} + 1}] = x^{3} + x$ . Therefore, $I = \int_{0}^{2} (x^{3} + x) d x = {[\frac{x^{4}}{4} + \frac{x^{2}}{2}]}_{0}^{2} = \frac{16}{4} + \frac{4}{2} = 4 + 2 = 6$ .

✓Verify

Check that the exponential part completely eliminates because $\frac{1}{1 + e^{u}} + \frac{1}{1 + e^{- u}} = 1$ . The integrand reduces strictly to the even part of the numerator integrated over $[0, 2]$ , yielding $\int_{0}^{2} (x^{3} + x) d x = 4 + 2 = 6$ .

Hints that build this answer step by step

What symmetry does the integrand $f (x) = \frac{∣ x^{3} + x ∣}{e^{∣ x ∣ ∣ x ∣} + 1}$ exhibit on $[- 2, 2]$ ?
$f (- x) = f (x)$ , so the function is even.
Using the even symmetry, how does the integral transform after substituting $t = x^{2}$ ?
$I = \int_{0}^{4} \frac{t + 1}{e^{t} + 1} d t$
What is the value of $\int_{0}^{4} \frac{t + 1}{e^{t} + 1} d t$ ?
$6$

✓ 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: 22 September 2026

Quick checks

Students also ask

Why does

1 / (e^{x^{2}} + 1) + 1 / (e^{- x^{2}} + 1)

equal 1?

Multiply the numerator and denominator of the second fraction by $e^{x^{2}}$ : $\frac{1}{e^{- x^{2}} + 1} = \frac{e^{x^{2}}}{1 + e^{x^{2}}}$ . Then $\frac{1 + e^{x^{2}}}{1 + e^{x^{2}}} = 1$ .

Answer

Because the integrand is an even function, the integral over $[- 2, 2]$ simplifies to $2 \int_{0}^{2} \frac{x^{3} + x}{e^{x^{2}} + 1} d x$ , which evaluates to $6$ after applying the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ or direct substitution.

Why each option works or fails

A: 5e^2 - Believing that integrating the exponential in the denominator leaves an unevaluated term proportional to $e^{2}$ . Notice that multiplying or pairing terms using King's property or substitution cancels out the exponential term entirely, leaving a simple polynomial.
B: 3e^(-2) - Assuming that the denominator $e^{x^{2}} + 1$ introduces negative exponential factors into the final value. Substitute $u = x^{2}$ to get $d u = 2 x d x$ , which transforms the integrand into $\frac{u + 1}{e^{u} + 1}$ , where the exponential disappears after using reflection symmetry or paired integrals.
C: 4 - Integrating without multiplying by $2$ when reducing the integral of the even function from $[- 2, 2]$ to $[0, 2]$ , or incorrectly evaluating ${[\frac{u^{2}}{2} + u]}_{0}^{4}$ . Ensure the factor of $2$ from the symmetric interval property $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ is properly included throughout the calculation.
D · correct: 6 - The integral is evaluated correctly by exploiting symmetry. The integrand is even, so the integral equals $2 \int_{0}^{2} \frac{x (x^{2} + 1)}{e^{x^{2}} + 1} d x$ . Letting $t = x^{2}$ gives $\int_{0}^{4} \frac{t + 1}{e^{t} + 1} d t$ . Using the reflection property $\int_{0}^{a} f (t) d t = \int_{0}^{a} f (a - t) d t$ cancels the denominator and yields $6$ .

Step-by-step solution

Given: The integral is $I = \int_{- 2}^{2} \frac{∣ x^{3} + x ∣}{e^{x ∣ x ∣} + 1} d x$ (noting the standard typo in the denominator of the stem where $e^{∣ x ∣ ∣ x ∣}$ is actually $e^{x ∣ x ∣}$ , consistent with standard JEE papers).
Goal: Evaluate $I = \int_{- 2}^{2} f (x) d x$ where $f (x) = \frac{∣ x^{3} + x ∣}{e^{x ∣ x ∣} + 1}$ .
Approach: Use King's property for symmetric limits $\int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x$ . Notice that the numerator $∣ x^{3} + x ∣$ is an even function, while the exponent $x ∣ x ∣$ in the denominator is an odd function.
Execute: For $x \in [0, 2]$ , we have $∣ x^{3} + x ∣ = x^{3} + x$ and $x ∣ x ∣ = x^{2}$ . Hence, $f (x) = \frac{x^{3} + x}{e^{x^{2}} + 1}$ . For $f (- x)$ , the numerator is $∣ (- x)^{3} + (- x) ∣ = ∣ - (x^{3} + x) ∣ = x^{3} + x$ , and the exponent is $(- x) ∣ - x ∣ = - x ∣ x ∣ = - x^{2}$ . Thus, $f (- x) = \frac{x^{3} + x}{e^{- x^{2}} + 1} = \frac{(x^{3} + x) e^{x^{2}}}{1 + e^{x^{2}}}$ .
Execute: Add $f (x)$ and $f (- x)$ on $[0, 2]$ : $f (x) + f (- x) = (x^{3} + x) [\frac{1}{e^{x^{2}} + 1} + \frac{e^{x^{2}}}{e^{x^{2}} + 1}] = x^{3} + x$ . Therefore, $I = \int_{0}^{2} (x^{3} + x) d x = {[\frac{x^{4}}{4} + \frac{x^{2}}{2}]}_{0}^{2} = \frac{16}{4} + \frac{4}{2} = 4 + 2 = 6$ .
Verify: Check that the exponential part completely eliminates because $\frac{1}{1 + e^{u}} + \frac{1}{1 + e^{- u}} = 1$ . The integrand reduces strictly to the even part of the numerator integrated over $[0, 2]$ , yielding $\int_{0}^{2} (x^{3} + x) d x = 4 + 2 = 6$ .

Definite integral of an even function involving absolute values

What feels right?

Because the integrand is an even function, the integral over $[- 2, 2]$ simplifies to $2 \int_{0}^{2} \frac{x^{3} + x}{e^{x^{2}} + 1} d x$ , which evaluates to $6$ after applying the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ or direct substitution.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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