Key Concepts and Formulas

• Absolute Value Function: The definition of $|x|$ is crucial: $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$. When integrating a function involving $|x|$ over an interval containing $0$, the integral must be split at $x=0$.
• Integration of Exponential Functions: The general formula for integrating an exponential function is $\int e^{ax} dx = \frac{e^{ax}}{a} + C$.
• Definite Integral Evaluation: The value of a definite integral $\int_a^b f(x) dx$ is found by evaluating the antiderivative $F(x)$ at the upper and lower limits: $F(b) - F(a)$.
• Logarithm Properties: Key properties include $\ln(e^k) = k$ and $\ln(a/b) = \ln a - \ln b$, which can be used to simplify logarithmic expressions. Also, $e^x > 0$ for all real $x$.

Step-by-Step Solution

We are asked to find the value of $\alpha$ for which $4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$.

Step 1: Handling the Absolute Value in the Integral The integrand contains $|x|$. Since the interval of integration is $[-1, 2]$, which includes $0$, we must split the integral at $x=0$. For $x \in [-1, 0)$, $|x| = -x$. So, $e^{-\alpha|x|} = e^{-\alpha(-x)} = e^{\alpha x}$. For $x \in [0, 2]$, $|x| = x$. So, $e^{-\alpha|x|} = e^{-\alpha x}$. Therefore, the integral can be rewritten as: $\int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + \int\limits_0^2 {{e^{ - \alpha x}}dx}$ The given equation becomes: $4\alpha \left( \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + \int\limits_0^2 {{e^{ - \alpha x}}dx} \right) = 5$

Step 2: Evaluating the Integrals We apply the integration formula $\int e^{ax} dx = \frac{e^{ax}}{a}$.

For the first integral, $\int\limits_{ - 1}^0 {{e^{\alpha x}}dx}$: $\left[ \frac{e^{\alpha x}}{\alpha} \right]_{ - 1}^0 = \frac{e^{\alpha(0)}}{\alpha} - \frac{e^{\alpha(-1)}}{\alpha} = \frac{1}{\alpha} - \frac{e^{-\alpha}}{\alpha} = \frac{1 - e^{-\alpha}}{\alpha}$

For the second integral, $\int\limits_0^2 {{e^{ - \alpha x}}dx}$: $\left[ \frac{e^{-\alpha x}}{-\alpha} \right]_0^2 = \frac{e^{-\alpha(2)}}{-\alpha} - \frac{e^{-\alpha(0)}}{-\alpha} = \frac{e^{-2\alpha}}{-\alpha} - \frac{1}{-\alpha} = \frac{1 - e^{-2\alpha}}{\alpha}$

Substitute these results back into the main equation: $4\alpha \left( \frac{1 - e^{-\alpha}}{\alpha} + \frac{1 - e^{-2\alpha}}{\alpha} \right) = 5$ We can factor out $\frac{1}{\alpha}$ from the terms inside the parenthesis: $4\alpha \cdot \frac{1}{\alpha} \left( (1 - e^{-\alpha}) + (1 - e^{-2\alpha}) \right) = 5$ The $\alpha$ terms cancel out: $4 \left( 1 - e^{-\alpha} + 1 - e^{-2\alpha} \right) = 5$

Step 3: Simplifying and Forming a Quadratic Equation Simplify the expression inside the parenthesis: $4 \left( 2 - e^{-\alpha} - e^{-2\alpha} \right) = 5$ Distribute the $4$: $8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5$ Rearrange the terms to form a quadratic equation in terms of $e^{-\alpha}$: $4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$

Step 4: Solving the Quadratic Equation Let $t = e^{-\alpha}$. Then $e^{-2\alpha} = (e^{-\alpha})^2 = t^2$. Substituting this into the equation gives: $4t^2 + 4t - 3 = 0$ We can solve this quadratic equation by factoring. We look for two numbers that multiply to $4 \times (-3) = -12$ and add to $4$. These numbers are $6$ and $-2$. $4t^2 + 6t - 2t - 3 = 0$ Factor by grouping: $2t(2t + 3) - 1(2t + 3) = 0$ $(2t - 1)(2t + 3) = 0$ This yields two possible solutions for $t$: $2t - 1 = 0 \Rightarrow t = \frac{1}{2}$ $2t + 3 = 0 \Rightarrow t = -\frac{3}{2}$

Step 5: Determining the Valid Solution for $\alpha$ Recall that we defined $t = e^{-\alpha}$. Since the exponential function $e^x$ is always positive for any real number $x$, $t$ must be positive. Therefore, $t = -\frac{3}{2}$ is an extraneous solution and must be rejected. The valid solution is $t = \frac{1}{2}$. Substitute back $t = e^{-\alpha}$: $e^{-\alpha} = \frac{1}{2}$ To solve for $\alpha$, take the natural logarithm of both sides: $\ln(e^{-\alpha}) = \ln\left(\frac{1}{2}\right)$ Using the property $\ln(e^k) = k$: $-\alpha = \ln\left(\frac{1}{2}\right)$ Using the property $\ln(1/b) = -\ln b$: $-\alpha = -\ln(2)$ Multiply by $-1$: $\alpha = \ln(2)$ This can also be written as $\alpha = \log_e 2$.

Common Mistakes & Tips

• Absolute Value Splitting: Failing to split the integral at $x=0$ when $|x|$ is involved is a common oversight.
• Sign Errors: Be extremely careful with negative signs during integration and when evaluating definite integrals, especially with terms like $e^{-\alpha x}$.
• Extraneous Solutions: Always check the validity of solutions obtained from algebraic equations, particularly when dealing with exponential functions where the base is positive.

Summary

The problem requires careful handling of an integral involving an absolute value function. By splitting the integral at $x=0$, integrating the exponential terms, and simplifying the resulting equation, we obtained a quadratic equation in terms of $e^{-\alpha}$. Solving this quadratic equation and considering the domain of the exponential function allowed us to find the unique valid value for $\alpha$.

The final answer is $\boxed{\log_e 2}$.