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Integral Calculus: JEE Main Mathematics Question with Solution

If the area of the region

\{(x, y) : -1 \le x \le 1, 0 \le y \le a + e^{|x|} - e^{-x}, a > 0\}

\frac{e^2 + 8e + 1}{e}

, then the value of

a

is :

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Step-by-step solutionView

Correct answer

Integrating the function $y = a + e^{|x|} - e^{-x}$ from $-1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .

Option analysis

Why each option works or fails

A · 8

Believing that $2a = 8 + e - e$ directly equals $8$ without dividing by $2$ , or equating $a$ to the numerator's coefficient of $e$ . After integrating, fully simplify the linear equation in terms of $a$ : isolate $2a = 10$ , which gives $a = 5$ rather than $8$ .

B · 7

Dropping the integration constant contributions from $e^{-x}$ or making a sign error in $\int_{-1}^1 e^{-x}\,dx = e - \frac{1}{e}$ . Evaluate $\int_{-1}^1 -e^{-x}\,dx = [e^{-x}]_{-1}^1 = e^{-1} - e^1 = \frac{1}{e} - e$ carefully, keeping track of negative signs.

C · 5

None. The integral of $a + e^{|x|} - e^{-x}$ over $[-1, 1]$ correctly simplifies to $2a + e - 1 - \frac{1}{e} + (e - 1) + 1 - e^{-1} = 2a + 2e - 2 - (e - e^{-1})$ , which equates to the given area to yield $a = 5$ . This is the correct result.

D · 6

Making an arithmetic error when combining the constant terms from $[e^x]$ and $[e^{-x}]$ limits, leading to $2a = 12$ instead of $2a = 10$ . Ensure that the constant terms from $(e^0 - e^{-1}) + (e^1 - e^0) = e - e^{-1}$ and other bounds are added precisely.

Reviewed route

Solution

StepWorking

01given

Region is bounded by $-1 \le x \le 1$ and $0 \le y \le a + e^{|x|} - e^{-x}$ , where $a > 0$ , and the area is $\frac{e^2 + 8e + 1}{e}$ .

02approach

Set up the definite integral $\text{Area} = \int_{-1}^1 (a + e^{|x|} - e^{-x})\,dx$ . Split the integral at $x = 0$ to handle $|x|$ .

03execute

Split into $[-1, 0]$ and $[0, 1]$ : $\text{Area} = \int_{-1}^0 (a + e^{-x} - e^{-x})\,dx + \int_0^1 (a + e^x - e^{-x})\,dx$ $= \int_{-1}^0 a\,dx + \int_0^1 (a + e^x - e^{-x})\,dx$ $= a(0 - (-1)) + [ax + e^x + e^{-x}]_0^1$ $= a + (a(1) + e^1 + e^{-1}) - (0 + e^0 + e^0)$ $= a + a + e + \frac{1}{e} - 2 = 2a + e + \frac{1}{e} - 2 = 2a - 2 + \frac{e^2 + 1}{e}$ .

04execute

Equate the computed area to the given value: $2a - 2 + \frac{e^2 + 1}{e} = \frac{e^2 + 8e + 1}{e} = \frac{e^2 + 1}{e} + 8$ $2a - 2 = 8 \implies 2a = 10 \implies a = 5$ .

✓verify

For $a=5$ , check integrand positivity: $5 + e^{|x|} - e^{-x} \ge 5 + 1 - e > 0$ for all $x \in [-1, 1]$ . Area $= 2(5) - 2 + \frac{e^2+1}{e} = 8 + \frac{e^2+1}{e} = \frac{e^2+8e+1}{e}$ , which matches exactly.

✓ 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 does the term

e^{-x} - e^{-x}

become 0 on

[-1, 0]

For $x \in [-1, 0]$ , $|x| = -x$ , so $e^{|x|} = e^{-x}$ . Thus $e^{|x|} - e^{-x} = e^{-x} - e^{-x} = 0$ .

Answer

Integrating the function $y = a + e^{|x|} - e^{-x}$ from $-1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .

Why each option works or fails

A: 8 - Believing that $2a = 8 + e - e$ directly equals $8$ without dividing by $2$ , or equating $a$ to the numerator's coefficient of $e$ . After integrating, fully simplify the linear equation in terms of $a$ : isolate $2a = 10$ , which gives $a = 5$ rather than $8$ .
B: 7 - Dropping the integration constant contributions from $e^{-x}$ or making a sign error in $\int_{-1}^1 e^{-x}\,dx = e - \frac{1}{e}$ . Evaluate $\int_{-1}^1 -e^{-x}\,dx = [e^{-x}]_{-1}^1 = e^{-1} - e^1 = \frac{1}{e} - e$ carefully, keeping track of negative signs.
C · correct: 5 - None. The integral of $a + e^{|x|} - e^{-x}$ over $[-1, 1]$ correctly simplifies to $2a + e - 1 - \frac{1}{e} + (e - 1) + 1 - e^{-1} = 2a + 2e - 2 - (e - e^{-1})$ , which equates to the given area to yield $a = 5$ . This is the correct result.
D: 6 - Making an arithmetic error when combining the constant terms from $[e^x]$ and $-1$ limits, leading to $2a = 12$ instead of $2a = 10$ . Ensure that the constant terms from $(e^0 - e^{-1}) + (e^1 - e^0) = e - e^{-1}$ and other bounds are added precisely.

Step-by-step solution

given: Region is bounded by $-1 \le x \le 1$ and $0 \le y \le a + e^{|x|} - e^{-x}$ , where $a > 0$ , and the area is $\frac{e^2 + 8e + 1}{e}$ .
approach: Set up the definite integral $\text{Area} = \int_{-1}^1 (a + e^{|x|} - e^{-x})\,dx$ . Split the integral at $x = 0$ to handle $|x|$ .
execute: Split into $[-1, 0]$ and $[0, 1]$ : $\text{Area} = \int_{-1}^0 (a + e^{-x} - e^{-x})\,dx + \int_0^1 (a + e^x - e^{-x})\,dx$ $= \int_{-1}^0 a\,dx + \int_0^1 (a + e^x - e^{-x})\,dx$ $= a(0 - (-1)) + [ax + e^x + e^{-x}]_0^1$ $= a + (a(1) + e^1 + e^{-1}) - (0 + e^0 + e^0)$ $= a + a + e + \frac{1}{e} - 2 = 2a + e + \frac{1}{e} - 2 = 2a - 2 + \frac{e^2 + 1}{e}$ .
execute: Equate the computed area to the given value: $2a - 2 + \frac{e^2 + 1}{e} = \frac{e^2 + 8e + 1}{e} = \frac{e^2 + 1}{e} + 8$ $2a - 2 = 8 \implies 2a = 10 \implies a = 5$ .
verify: For $a=5$ , check integrand positivity: $5 + e^{|x|} - e^{-x} \ge 5 + 1 - e > 0$ for all $x \in [-1, 1]$ . Area $= 2(5) - 2 + \frac{e^2+1}{e} = 8 + \frac{e^2+1}{e} = \frac{e^2+8e+1}{e}$ , which matches exactly.

Integral Calculus: JEE Main Mathematics Question with Solution

Integrating the function y=a+e∣x∣−e−xy = a + e^{|x|} - e^{-x}y=a+e∣x∣−e−x from −1-1−1 to 111 gives the area 2a + 2e - 1 - rac{1}{e}; equating this to rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e} yields a=5a = 5a=5.

01given

02approach

03execute

04execute

✓verify

Students also ask

Integrating the function $y = a + e^{|x|} - e^{-x}$ from $-1$ to $1$ gives the area $2a + 2e - 1 - rac{1}{e}$ ; equating this to $rac{e^2 + 8e + 1}{e} = e + 8 + rac{1}{e}$ yields $a = 5$ .