Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the curves $y=e^x$, $y=|e^x-1|$ and y-axis is:

Welcome, fellow explorer of the mathematical landscape. Today, we are mapping a territory defined by three boundaries: the exponential growth of $y=e^x$, the reflective nature of the modulus function $y=|e^x-1|$, and the vertical wall of the y-axis, $x=0$.

To find the area trapped between them, we must first understand the "personality" of these functions.

Imagine the function $y=|e^x-1|$ as a mirror. For $x\ge 0$, $e^x\ge 1$, so the modulus does nothing—it remains $e^x-1$.

For $x<0$, the value $e^x-1$ is negative. The modulus acts like a mirror, reflecting that negative portion across the x-axis to become $1-e^x$. This is the crucial realization: the behavior of our boundary changes exactly at the y-axis.

Now, we hunt for the intersection by setting $e^x=|e^x-1|$. As discussed, for $x>0$, there is no intersection.

For $x<0$, we solve the equation $e^x=1-e^x$. This leads us to:

Taking the natural logarithm of both sides, we find our boundary point: $x=\ln (\frac{1}{2})$, which is $-\ln 2$. We have successfully defined our limits of integration: from $-\ln 2$ to $0$.

With our limits set, we construct the integral for the area $A$. The area is the accumulation of the vertical distance between the upper curve $y=e^x$ and the lower curve $y=1-e^x$.

The integral becomes:

Look at how the expression simplifies. The negative sign distributes, and we are left with:

This is the moment of truth. We integrate term by term, where the integral of $2e^x$ is $2e^x$ and the integral of $1$ is $x$. We are left with the evaluation:

Now, we apply the Fundamental Theorem of Calculus. At the upper limit $x=0$, we have:

At the lower limit $x=-\ln 2$, we have $2e^{-\ln 2}-(-\ln 2)$. Since $e^{-\ln 2}=\frac{1}{e^{\ln 2}}=\frac{1}{2}$, this simplifies to:

Subtracting the lower limit value from the upper limit value, we get:

And there it is! The final area is $1-\ln 2$. Remember, the modulus function is not an obstacle; it is a signpost telling you where to split your journey.