1. Understand the curves

We are given:

• $y=e^x$
• $y=|e^x-1|$
• the $y$-axis, i.e. $x=0$

We first simplify $|e^x-1|$ depending on $x$.

Since $e^x<1$ for $x<0$, we get $|e^x-1|=1-e^x \qquad (x<0).$

Since $e^x\ge 1$ for $x\ge 0$, we get $|e^x-1|=e^x-1 \qquad (x\ge 0).$

Because the region is enclosed with the $y$-axis ($x=0$), the enclosed part must lie to the left of the $y$-axis, i.e. for $x\le 0$.

So in the relevant region, the curves are:

• upper curve: $y=e^x$
• lower curve: $y=1-e^x$

2. Find their point of intersection

Set $e^x=1-e^x.$ So, $2e^x=1$ $e^x=\frac12$ $x=\ln\left(\frac12\right)=-\ln 2.$

Thus the two curves intersect at $x=-\ln 2.$

At $x=0$:

• $y=e^0=1$
• $y=|e^0-1|=0$

Hence the region enclosed by the two curves and the line $x=0$ lies for $-\ln 2 \le x \le 0.$

3. Set up the area integral

Area $A=\int_{-\ln 2}^{0} \left[e^x-(1-e^x)\right]dx.$

Simplify the integrand: $e^x-(1-e^x)=2e^x-1.$

So, $A=\int_{-\ln 2}^{0} (2e^x-1)\,dx.$

4. Evaluate the integral

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

Therefore, $A=\left[2e^x-x\right]_{-\ln 2}^{0}.$

At $x=0$: $2e^0-0=2.$

At $x=-\ln 2$: $2e^{-\ln 2}-(-\ln 2)=2\cdot \frac12+\ln 2=1+\ln 2.$

Hence, $A=2-(1+\ln 2)=1-\ln 2.$

5. Match with options

$A=1-\ln 2$ which corresponds to Option C.