JEE Main 2025 Jan 24 Shift 2, Mathematics Q17: Area as Definite Integral

JEE Main2025Jan 24, Shift 2Mathematics

The area of the region enclosed by the curves $y = e^{x}$ , $y = ∣ e^{x} - 1∣$ and the y-axis is:

A
$1 + lo g_{e} 2$
B
$lo g_{e} 2$
C
$2 lo g_{e} 2 - 1$
D
$1 - lo g_{e} 2$

Solution

Note that $∣ e^{x} - 1∣$ equals $e^{x} - 1$ for $x \geq 0$ and $1 - e^{x}$ for $x < 0$ .

For $x \geq 0$ : the two curves $y = e^{x}$ and $y = e^{x} - 1$ are parallel and never meet, so the region they bound together with the $y$ -axis is unbounded there.

The enclosed region therefore lies in $x < 0$ , where $∣ e^{x} - 1∣ = 1 - e^{x}$ . The upper curve is $y = e^{x}$ and the lower curve is $y = 1 - e^{x}$ (for $x < 0$ , $1 - e^{x} > 0$ ). They intersect where $e^{x} = 1 - e^{x} \Rightarrow e^{x} = 1/2 \Rightarrow x = - ln 2$ .

The region is bounded on the right by the $y$ -axis ( $x = 0$ ) and on the left by the intersection point $x = - ln 2$ .

$Area = \int_{- l n 2}^{0} [e^{x} - (1 - e^{x})] d x = \int_{- l n 2}^{0} (2 e^{x} - 1) d x$

$= [2 e^{x} - x]_{- l n 2}^{0} = (2 - 0) - (2 \cdot \frac{1}{2} + ln 2) = 2 - 1 - ln 2 = 1 - ln 2$

Correct option: (4)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q17: Area as Definite Integral

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