Question

If the area of the region \[ \{(x, y) : -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a> 0\} \] is \[ \frac{e^2 + 8e + 1}{e}, \] then the value of \(a\) is:

• A. 7
• B. 6
• C. 8
• D. 5

Hint:

When dealing with absolute values in integrals, split the integral into regions where the absolute value expression simplifies. This makes the problem easier to handle.

Answer 1

The Correct Option is D

The area of the region is given by the integral over the specified region: \[ \text{Area} = \int_{-1}^{1} \left( a + e^{|x|} - e^{-x} \right) \, dx. \] We can split the integral into two parts based on the absolute value function \( e^{|x|} \). For \( x \geq 0 \), \( e^{|x|} = e^x \), and for \( x <0 \), \( e^{|x|} = e^{-x} \). Thus, the area becomes: \[ \text{Area} = \int_{-1}^{0} \left( a + e^{-x} - e^{-x} \right) \, dx + \int_{0}^{1} \left( a + e^{x} - e^{-x} \right) \, dx. \] Simplifying each integral, we get: \[ \text{Area} = \int_{-1}^{0} a \, dx + \int_{0}^{1} a \, dx + \int_{0}^{1} e^x \, dx - \int_{0}^{1} e^{-x} \, dx. \] The first two integrals are straightforward: \[ \int_{-1}^{0} a \, dx = a, \quad \int_{0}^{1} a \, dx = a. \] Now we calculate the exponential integrals: \[ \int_{0}^{1} e^x \, dx = e - 1, \quad \int_{0}^{1} e^{-x} \, dx = 1 - \frac{1}{e}. \] Thus, the area is: \[ \text{Area} = 2a + (e - 1) - (1 - \frac{1}{e}). \] Simplifying this expression: \[ \text{Area} = 2a + e - 1 - 1 + \frac{1}{e} = 2a + e - 2 + \frac{1}{e}. \] We are given that the area is \( \frac{e^2 + 8e + 1}{e} \). Equating this with the expression for the area, we get: \[ 2a + e - 2 + \frac{1}{e} = \frac{e^2 + 8e + 1}{e}. \] Multiplying both sides by \( e \) to eliminate the denominator: \[ e(2a + e - 2 + \frac{1}{e}) = e^2 + 8e + 1, \] \[ e(2a) + e^2 - 2e + 1 = e^2 + 8e + 1. \] Simplifying: \[ 2ae + e^2 - 2e + 1 = e^2 + 8e + 1. \] Canceling out \( e^2 + 1 \) from both sides: \[ 2ae - 2e = 8e. \] Factoring out \( e \): \[ e(2a - 2) = 8e. \] Dividing both sides by \( e \): \[ 2a - 2 = 8. \] Solving for \( a \): \[ 2a = 10 \quad \Rightarrow \quad a = 5. \] Conclusion: The correct answer is (4), as \( a = 5 \).