Question

Area bounded by the curves \( y = e^x \), \( y = e^{-x} \) and the straight line \( x = 1 \) is (in sq. units):

• A. \( e + \frac{1}{e} \)
• B. \( e + \frac{1}{e} + 2 \)
• C. \( e + \frac{1}{e} - 2 \)
• D. \( e - \frac{1}{e} + 2 \)
• E. \( e - \frac{1}{e} \)

Hint:

Drawing a quick sketch of $e^x$ and $e^{-x}$ helps verify which function is the "upper" one in the integral. $e^x$ grows while $e^{-x}$ decays, so for positive $x$, $e^x$ is always higher.

Answer 1

The Correct Option is C

Concept: The area between two curves \( f(x) \) and \( g(x) \) from \( x = a \) to \( x = b \) is given by the integral: \[ \text{Area} = \int_{a}^{b} |f(x) - g(x)| \, dx \] First, we find the intersection point of \( y = e^x \) and \( y = e^{-x} \) to determine the lower limit.

Step 1: Find the point of intersection.
Set \( e^x = e^{-x} \): \[ e^{2x} = 1 \quad \Rightarrow \quad 2x = 0 \quad \Rightarrow \quad x = 0 \] At \( x = 0 \), the curves intersect at \( (0, 1) \). For the interval \( x \in [0, 1] \), \( e^x \ge e^{-x} \).

Step 2: Set up the definite integral.
The area is bounded from \( x = 0 \) to \( x = 1 \): \[ \text{Area} = \int_{0}^{1} (e^x - e^{-x}) \, dx \]

Step 3: Integrate and evaluate.
\[ \text{Area} = [e^x - (-e^{-x})]_{0}^{1} = [e^x + e^{-x}]_{0}^{1} \] \[ = (e^1 + e^{-1}) - (e^0 + e^0) \] \[ = e + \frac{1}{e} - (1 + 1) = e + \frac{1}{e} - 2 \]