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

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Area Under Curves

The area of the region enclosed by the curves

y = e^{x}, y = ∣ e^{x} - 1 ∣

and

y

-axis is:

Answer: (D)

The curve

y = ∣ e^{x} - 1∣

equals

e^{x} - 1

for

x \geq 0

and

1 - e^{x}

for

x < 0

. Since

y = e^{x}

is always above

e^{x} - 1

(they never meet for

x \geq 0

), the curves

y = e^{x}

and

y = ∣ e^{x} - 1∣

can only intersect where

x < 0

e^{x} = 1 - e^{x} \Rightarrow 2 e^{x} = 1 \Rightarrow x = - ln 2.

For

- ln 2 < x < 0

e^{x} > 1 - e^{x}

(check

x = - ln 2/2

e^{x} \approx 0.71

1 - e^{x} \approx 0.29

), so the region enclosed by

y = e^{x}

(top),

y = ∣ e^{x} - 1∣ = 1 - e^{x}

(bottom), and the

y

-axis (

x = 0

, right boundary) runs from

x = - ln 2

x = 0

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} .

Evaluating:

= (2 \cdot 1 - 0) - (2 \cdot e^{- l n 2} - (- ln 2)) = 2 - (2 \cdot \frac{1}{2} + ln 2) = 2 - (1 + ln 2) = 1 - ln 2.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option