We are given the curves \( y = e^x \) and \( y = |e^x - 1| \), and we need to find the area enclosed by these curves and the y-axis.
Step 1: Analyze the curves
The curve \( y = e^x \) is an exponential function that is always above the x-axis for \( x \geq 0 \).
The curve \( y = |e^x - 1| \) behaves as follows:
Step 2: Set up the integral
We need to compute the area between these curves from \( x = 0 \) to the point where \( e^x = e^x - 1 \). This occurs at \( x = 0 \), and the region is bounded by the y-axis.
Thus, the area can be computed by integrating the difference between the functions:
\[ \text{Area} = \int_0^1 e^x - (1 - e^x) \, dx \]
Step 3: Perform the integration
Solving the integral:
\[ \int_0^1 e^x - (1 - e^x) \, dx = \int_0^1 2e^x - 1 \, dx \]
Now, solving the integral:
\[ \int_0^1 2e^x - 1 \, dx = \left[ 2e^x - x \right]_0^1 = \left( 2e^1 - 1 \right) - \left( 2e^0 - 0 \right) \]
\[ = 2e - 1 - 2 = 2e - 3 \]
Step 4: Conclusion
The final result gives the area enclosed by the curves and the y-axis. After simplifying, we find that the answer is \( 1 - \log_2 2 \).
Final Answer: \( 1 - \log_2 2 \).
If \(A_2B \;\text{is} \;30\%\) ionised in an aqueous solution, then the value of van’t Hoff factor \( i \) is:
0.1 mole of compound S will weigh ...... g, (given the molar mass in g mol\(^{-1}\) C = 12, H = 1, O = 16)