JEE Main 2025 Jan 23 Shift 2, Mathematics Q8: Area as Definite Integral

JEE Main2025Jan 23, Shift 2Mathematics

If the area of the region ${(x, y) : - 1 \leq x \leq 1, 0 \leq y \leq a + e^{∣ x ∣} - e^{- x}, a > 0}$ is $\frac{e ^{2} + 8 e + 1}{e}$ , then the value of $a$ is :

Solution

Split the integral at $x = 0$ to handle $∣ x ∣$ .

For $x \in [- 1, 0]$ : $e^{∣ x ∣} - e^{- x} = e^{- x} - e^{- x} = 0$ , so the integrand is just $a$ .

For $x \in [0, 1]$ : $e^{∣ x ∣} - e^{- x} = e^{x} - e^{- x}$ , so the integrand is $a + e^{x} - e^{- x}$ .

Area $= \int_{- 1}^{0} a d x + \int_{0}^{1} (a + e^{x} - e^{- x}) d x = a + [a x + e^{x} + e^{- x}]_{0}^{1} = a + (a + e + e^{- 1} - 2) = 2 a + e + \frac{1}{e} - 2.$

Set equal to $\frac{e ^{2} + 8 e + 1}{e} = e + 8 + \frac{1}{e}$ :

$2 a + e + \frac{1}{e} - 2 = e + 8 + \frac{1}{e} \Rightarrow 2 a = 10 \Rightarrow a = 5.$

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q8: Area as Definite Integral

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