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

To find the value of \(a\), we first calculate the area of the region defined by the set: \( \{(x, y) : -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a>0 \} \).

The area is obtained by integrating the upper boundary curve over the specified \(x\) interval:

\(Area = \int_{-1}^{1} (a + e^{|x|} - e^{-x}) \, dx\)

Due to the absolute value function, we split the integral at \(x = 0\):

\(Area = \int_{-1}^{0} (a + e^{-x} - e^{-x}) \, dx + \int_{0}^{1} (a + e^{x} - e^{-x}) \, dx\)

Simplify the integrals:

1. For \(x \in [-1, 0]\), \(e^{|x|} = e^{-x}\). The integral becomes:

\(\int_{-1}^{0} (a + e^{-x} - e^{-x}) \, dx = \int_{-1}^{0} a \, dx = a \cdot (0 - (-1)) = a\)

2. For \(x \in [0, 1]\), \(e^{|x|} = e^{x}\). The integral is:

\(\int_{0}^{1} (a + e^{x} - e^{-x}) \, dx = \int_{0}^{1} (a + e^{x} - e^{-x}) \, dx = a + [e^{x} - e^{-x}]_{0}^{1}\)

Evaluate the antiderivative at the limits:

\([e^{x} - e^{-x}]_{0}^{1} = (e^{1} - e^{-1}) - (e^{0} - e^{0}) = e - \frac{1}{e}\)

The area from \(x = 0\) to \(x = 1\) is \(a + e - \frac{1}{e}\).

The total area is the sum of the areas from the two intervals: \(a + (a + e - \frac{1}{e}) = 2a + e - \frac{1}{e}\).

We are given that the total area is \( \frac{e^2 + 8e + 1}{e} \). Thus:

\(2a + e - \frac{1}{e} = \frac{e^2 + 8e + 1}{e}\)

To solve for \(a\), multiply both sides by \(e\):

\(2ae + e(e) - e(\frac{1}{e}) = e^2 + 8e + 1\)

\(2ae + e^2 - 1 = e^2 + 8e + 1\)

Cancel \(e^2\) from both sides and rearrange:

\(2ae = 8e + 2\)

Divide by \(2e\) to isolate \(a\):

\(a = \frac{8e + 2}{2e}\)

\(a = \frac{8e}{2e} + \frac{2}{2e}\)

\(a = 4 + \frac{1}{e}\)

The value of \(a\) is \(4 + \frac{1}{e}\).