### Related Formula The area between two curves y_1(x)$y_1(x)$ and y_2(x)$y_2(x)$ from x = a$x = a$ to x = b$x = b$ is given by: textArea = int_a^b |y_1(x) - y_2(x)| dx$\text{Area} = \int_{a}^{b} |y_1(x) - y_2(x)| dx$ ### Core Logic Analyze the curves to locate intersection points : - Curve 1: y = e^x$y = e^x$ - Curve 2: y = |e^x - 1|$y = |e^x - 1|$ - Boundary: y-axis (x = 0$x = 0$) For x < 0$x < 0$, e^x < 1 Rightarrow |e^x - 1| = 1 - e^x$e^x < 1 \Rightarrow |e^x - 1| = 1 - e^x$ . Find intersection: e^x = 1 - e^x Rightarrow 2e^x = 1 Rightarrow e^x = frac12 Rightarrow x = -ln 2$e^x = 1 - e^x \Rightarrow 2e^x = 1 \Rightarrow e^x = \frac{1}{2} \Rightarrow x = -\ln 2$.

### Step 1: Set up the Definite Integral The integration spans from x = -ln 2$x = -\ln 2$ to x = 0$x = 0$. In this interval, e^x ge 1 - e^x$e^x \ge 1 - e^x$: textArea = int_-ln 2^0 left[ e^x - (1 - e^x) right] dx$\text{Area} = \int_{-\ln 2}^{0} \left[ e^x - (1 - e^x) \right] dx$ textArea = int_-ln 2^0 (2e^x - 1) dx$\text{Area} = \int_{-\ln 2}^{0} (2e^x - 1) dx$ ### Step 2: Integration Evaluation Integrate term-by-term : textArea = left[ 2e^x - x right]_-ln 2^0$\text{Area} = \left[ 2e^x - x \right]_{-\ln 2}^{0}$ = left(2e^0 - 0right) - left(2e^-ln 2 - (-ln 2)right)$= \left(2e^0 - 0\right) - \left(2e^{-\ln 2} - (-\ln 2)\right)$ = 2 - left(2left(frac12right) + ln 2right) = 2 - (1 + ln 2) = 1 - ln 2$= 2 - \left(2\left(\frac{1}{2}\right) + \ln 2\right) = 2 - (1 + \ln 2) = 1 - \ln 2$ ### Pattern Recognition Modulus graphs split at their critical point (e^x = 1 Rightarrow x = 0$e^x = 1 \Rightarrow x = 0$). Recognizing that the required segment lies strictly in the negative quadrant simplifies the definition of the upper and lower functions immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals