Question

The value of $\int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx$ is equal to

• A. $3 - \frac{2\sqrt{2}}{3}$
• B. $2 + \frac{2\sqrt{2}}{3}$
• C. $1 - \frac{2\sqrt{2}}{3}$
• D. $1 + \frac{2\sqrt{2}}{3}$

Hint:

Simplify the integrand before evaluating the integral.

Answer 1

The Correct Option is D

1. Simplify the integrand: \[ \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \] \[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \] \[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
2. Evaluate the integral: \[ = \int_{-1}^{1} (1 + \sqrt{|x| - x}) \, dx \] \[ = \int_{-1}^{1} 1 \, dx + \int_{-1}^{1} \sqrt{|x| - x} \, dx \] \[ = [x]_{-1}^{1} + \int_{0}^{1} \sqrt{x} \, dx \] \[ = 2 + \frac{2\sqrt{2}}{3} \] Therefore, the correct answer is (4) $1 + \frac{2\sqrt{2}}{3}$.

Answer 2

We are to evaluate:

\[ I = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]

Concept Used:

Break the integral into regions depending on the sign of \(x\) because of \(|x|\). For \(x \ge 0\), \(|x| = x\); for \(x < 0\), \(|x| = -x\).

Step-by-Step Solution:

Step 1: Case 1 – for \(x \ge 0\):

\[ |x| - x = 0 \quad \Rightarrow \quad \sqrt{|x| - x} = 0 \] \[ \Rightarrow f(x) = \frac{(1 + 0)e^x + (0)e^{-x}}{e^x + e^{-x}} = \frac{e^x}{e^x + e^{-x}} \]

Step 2: Case 2 – for \(x < 0\):

\[ |x| = -x \Rightarrow |x| - x = -x - x = -2x \] \[ \sqrt{|x| - x} = \sqrt{-2x} \quad \text{(since }x<0\text{, } -2x>0) \] \[ \Rightarrow f(x) = \frac{(1 + \sqrt{-2x})e^x + (\sqrt{-2x})e^{-x}}{e^x + e^{-x}} \]

Step 3: Use symmetry to simplify the integral.

Let’s test \(f(-x)\) for \(x>0\):

\[ f(-x) = \frac{(1 + \sqrt{|{-x}| - (-x)})e^{-x} + (\sqrt{|{-x}| - (-x)})e^{x}}{e^{-x} + e^{x}} \] Since \(|-x| = x\), \[ |{-x}| - (-x) = x + x = 2x \Rightarrow \sqrt{|{-x}| - (-x)} = \sqrt{2x} \] \[ f(-x) = \frac{(1 + \sqrt{2x})e^{-x} + (\sqrt{2x})e^{x}}{e^x + e^{-x}} \]

Hence the integrand is different for positive and negative parts. We compute each separately.

Step 4: Split the integral:

\[ I = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx \] For \(x>0\): \[ f(x) = \frac{e^x}{e^x + e^{-x}} = \frac{1}{1 + e^{-2x}} \] For \(x<0\): \[ f(x) = \frac{(1 + \sqrt{-2x})e^x + (\sqrt{-2x})e^{-x}}{e^x + e^{-x}} \] This looks complex, so let’s use substitution symmetry: let \(x=-t\) in the first part, with \(t\in[0,1]\). Then: \[ I = \int_{0}^{1} f(-t)(-dt) + \int_{0}^{1} f(x)\,dx = \int_{0}^{1} [f(x) + f(-x)]\,dx \] \[ \Rightarrow I = \int_{0}^{1} \frac{e^x + (1+\sqrt{2x})e^{-x} + \sqrt{2x}e^{x}}{e^x + e^{-x}}\,dx \] Combine: \[ f(x)+f(-x) = \frac{(1+\sqrt{2x})(e^x + e^{-x})}{e^x + e^{-x}} = 1+\sqrt{2x} \] \[ \Rightarrow I = \int_0^1 (1 + \sqrt{2x})\,dx \]

Step 5: Evaluate the final simple integral.

\[ I = \int_0^1 1\,dx + \int_0^1 \sqrt{2x}\,dx = [x]_0^1 + \sqrt{2}\int_0^1 x^{1/2}\,dx \] \[ = 1 + \sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^1 = 1 + \frac{2\sqrt{2}}{3} \]

Final Computation & Result

The value of the given integral is:

\[ \boxed{1 + \frac{2\sqrt{2}}{3}} \]