Question:
The area of the figure bounded by \( y = e^x \), \( y = e^{-x} \) and the straight line \( x = 1 \) is:
The area of the figure bounded by \( y = e^x \), \( y = e^{-x} \) and the straight line \( x = 1 \) is:
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When calculating the area between two curves, set up the integral with the difference of the functions over the given limits.
Updated On: Apr 22, 2026
- \( \left( e + \frac{1}{e} \right) \, \text{sq unit} \)
- \( \left( e - \frac{1}{e} \right) \, \text{sq unit} \)
- \( \left( e + \frac{1}{e} - 2 \right) \, \text{sq unit} \)
- \( \left( e + \frac{1}{e} + 2 \right) \, \text{sq unit} \)
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The Correct Option is C
Solution and Explanation
Step 1: Set up the integral to find the area between the curves.
We need to calculate the area bounded by the curves \( y = e^x \), \( y = e^{-x} \), and the line \( x = 1 \). We will find the area between these two curves over the interval from 0 to 1. The total area is given by: \[ A = \int_0^1 \left( e^x - e^{-x} \right) \, dx \]
Step 2: Evaluate the integral.
We need to compute: \[ A = \int_0^1 e^x \, dx - \int_0^1 e^{-x} \, dx \] The integral of \( e^x \) is \( e^x \), and the integral of \( e^{-x} \) is \( -e^{-x} \).
Step 3: Apply the limits of integration.
Now, substitute the limits from 0 to 1: \[ A = \left[ e^x \right]_0^1 - \left[ -e^{-x} \right]_0^1 \] This gives: \[ A = \left( e^1 - e^0 \right) - \left( -e^{-1} + e^0 \right) \] \[ A = \left( e - 1 \right) - \left( -\frac{1}{e} + 1 \right) \]
Step 4: Simplify the expression.
Simplifying the expression: \[ A = e - 1 + \frac{1}{e} - 1 \] \[ A = e + \frac{1}{e} - 2 \]
Step 5: Conclusion.
Thus, the area of the figure is \( e + \frac{1}{e} - 2 \), which corresponds to option (C).
We need to calculate the area bounded by the curves \( y = e^x \), \( y = e^{-x} \), and the line \( x = 1 \). We will find the area between these two curves over the interval from 0 to 1. The total area is given by: \[ A = \int_0^1 \left( e^x - e^{-x} \right) \, dx \]
Step 2: Evaluate the integral.
We need to compute: \[ A = \int_0^1 e^x \, dx - \int_0^1 e^{-x} \, dx \] The integral of \( e^x \) is \( e^x \), and the integral of \( e^{-x} \) is \( -e^{-x} \).
Step 3: Apply the limits of integration.
Now, substitute the limits from 0 to 1: \[ A = \left[ e^x \right]_0^1 - \left[ -e^{-x} \right]_0^1 \] This gives: \[ A = \left( e^1 - e^0 \right) - \left( -e^{-1} + e^0 \right) \] \[ A = \left( e - 1 \right) - \left( -\frac{1}{e} + 1 \right) \]
Step 4: Simplify the expression.
Simplifying the expression: \[ A = e - 1 + \frac{1}{e} - 1 \] \[ A = e + \frac{1}{e} - 2 \]
Step 5: Conclusion.
Thus, the area of the figure is \( e + \frac{1}{e} - 2 \), which corresponds to option (C).
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