The area of the region enclosed by the curves y = e^x, y = |e^x -… — Solution | JEE Mathematics

MCQMediumJEE 2025Applications of Integrals (Area)

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

A
$1 + \log_2 2$
B
$\log_2 2$
C
$2 \log_2 2 - 1$
D
$1 - \log_2 2$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The curves are $y = e^x$ and $y = |e^x - 1|$ , and the region is enclosed with the y-axis.

Find: The enclosed area.

For $x < 0$ , we have $e^x - 1 < 0$ , so

y = |e^x - 1| = 1 - e^x

The intersection with $y = e^x$ is obtained from

e^x = 1 - e^x

2e^x = 1

and hence

e^x = \frac{1}{2}, \qquad x = -\log_e 2

Detailed Working from Extracted Solution

Between $x = -\log_e 2$ and $x = 0$ , the upper curve is $y = e^x$ and the lower curve is $y = 1 - e^x$ . Therefore,

\text{Area} = \int_{-\log_e 2}^{0} \left(e^x - (1 - e^x)\right) dx

which becomes

\text{Area} = \int_{-\log_e 2}^{0} (2e^x - 1) \, dx

Evaluate and Compare with Options

Now evaluate:

\int (2e^x - 1) \, dx = 2e^x - x

So,

\text{Area} = \left[2e^x - x\right]_{-\log_e 2}^{0}

= (2 - 0) - \left(2e^{-\log_e 2} + \log_e 2\right)

Since $e^{-\log_e 2} = \frac{1}{2}$ ,

\text{Area} = 2 - (1 + \log_e 2) = 1 - \log_e 2

The extracted solution declares the correct option as D, and the option text is $1 - \log_2 2$ . This notation is inconsistent because $\log_2 2 = 1$ , while the working gives $1 - \log_e 2$ . Following the recorded answer, the correct option is D.

Common mistakes

Using $|e^x - 1| = e^x - 1$ for all $x$ is wrong because the absolute value changes form when $x < 0$ . For the enclosed region to the left of the y-axis, use $|e^x - 1| = 1 - e^x$ .
Setting the limits as $0$ to $-\log_e 2$ without correcting orientation can produce a negative integral. Area must be taken with limits in increasing order, or the absolute value of the integral must be used.
Assuming the curves intersect at $x = 0$ is incorrect. At $x = 0$ , the values are $1$ and $0$ respectively, so they do not meet there. First solve $e^x = |e^x - 1|$ to get the actual intersection.

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Answer

Step-by-step solution

Common mistakes

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