Question

The value of $\alpha $ for which $4\alpha \int\limits_{-1}^{2}{{{e}^{-\alpha |x|}}dx=5}$ , is.

Answer 1

Hint: Start by simplification of the definite integral using the property $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)dx}$ to write \[\int\limits_{-1}^{2}{{{e}^{-\alpha |x|}}dx}=\int\limits_{-1}^{0}{{{e}^{-\alpha |x|}}dx}+\int\limits_{0}^{2}{{{e}^{-\alpha |x|}}dx}\] and use the fact that |x| opens with a positive sign in (0,2) and with negative sign in (-1,0). After simplification, use the formula $\int{{{e}^{kx}}}dx=\dfrac{{{e}^{kx}}}{k}+c$ .

Complete step-by-step answer:
Before starting the solution, let us discuss the important properties of definite integration.
Some important properties are:
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx+}\int\limits_{c}^{b}{f\left( x \right)dx}$
Now let us start with the integral given in the above question.
$4\alpha \int\limits_{-1}^{2}{{{e}^{-\alpha |x|}}dx=5}$
Using the property $\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)dx}$ , we get
\[4\alpha \int\limits_{-1}^{0}{{{e}^{-\alpha |x|}}dx}+4\alpha \int\limits_{0}^{2}{{{e}^{-\alpha |x|}}dx=5}\]
Now if we use the fact that |x| opens with a positive sign in (0,2) and with negative sign in (-1,0), we get
\[4\alpha \int\limits_{-1}^{0}{{{e}^{-\alpha \left( -x \right)}}dx}+4\alpha \int\limits_{0}^{2}{{{e}^{-\alpha x}}dx=5}\]
\[\Rightarrow 4\alpha \int\limits_{-1}^{0}{{{e}^{\alpha x}}dx}+4\alpha \int\limits_{0}^{2}{{{e}^{-\alpha x}}dx=5}\]
Now, if we use the formula $\int{{{e}^{kx}}}dx=\dfrac{{{e}^{kx}}}{k}+c$ , we get
\[4\alpha \left. \dfrac{{{e}^{\alpha x}}}{\alpha } \right|_{-1}^{0}+4\alpha \left. \dfrac{{{e}^{-\alpha x}}}{-\alpha } \right|_{0}^{2}=5\]
\[\Rightarrow 4\alpha \left( \dfrac{{{e}^{0}}-{{e}^{-\alpha }}}{\alpha } \right)-4\alpha \left( \dfrac{{{e}^{-2\alpha }}-{{e}^{0}}}{\alpha } \right)=5\]
\[\Rightarrow 4{{e}^{0}}-4{{e}^{-\alpha }}+4{{e}^{0}}-4{{e}^{-2\alpha }}=5\]
Now, we know that ${{e}^{0}}=1$ .
\[4-4{{e}^{-\alpha }}+4-4{{e}^{-2\alpha }}=5\]
\[\Rightarrow -4{{e}^{-\alpha }}-4{{e}^{-2\alpha }}=-3\]
\[\Rightarrow 4{{e}^{-2\alpha }}+4{{e}^{-\alpha }}-3=0\]
\[\Rightarrow 4{{e}^{-2\alpha }}+6{{e}^{-\alpha }}-2{{e}^{-\alpha }}-3=0\]
\[\Rightarrow \left( 2{{e}^{-\alpha }}-1 \right)\left( 2{{e}^{-\alpha }}+3 \right)=0\]
So, either \[\left( 2{{e}^{-\alpha }}-1 \right)=0\] else \[\left( 2{{e}^{-\alpha }}+3 \right)=0\] . But \[\left( 2{{e}^{-\alpha }}+3 \right)\] cannot be zero as e to the power some constant can never be negative.
\[\therefore 2{{e}^{-\alpha }}-1=0\]
\[{{e}^{-\alpha }}=\dfrac{1}{2}\]
Now we will take log of both sides.
\[\begin{align}
  & \log {{e}^{-\alpha }}=\log \dfrac{1}{2} \\
 & \Rightarrow -\alpha \log e=\log \dfrac{1}{2} \\
\end{align}\]
We know that loge=1 and $-\log a=\log \dfrac{1}{a}$ .
\[\alpha =-\log \dfrac{1}{2}\]
\[\Rightarrow \alpha =\log 2\]
Therefore, the answer to the above question is \[\alpha =\log 2.\]

Note: In case of questions related to definite integral the use of the right properties is very important. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well. Also, the function |x| can be defined as:
$|x|=\left\{ \begin{align}
  & -x\text{ , x0} \\
 & x\text{ , x}\ge \text{0} \\
\end{align} \right.$ .