Question

Find the value of \[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}\].

Answer 1

- Hint: To find the value of the given expression \[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}\], we will use the formula of integration of \[{{e}^{ax}}\]. The formula is given as \[\int\limits_{b}^{c}{{{e}^{ax}}dx}={{\left[ \dfrac{{{e}^{ax}}}{a} \right]}^{c}}_{b}+C\], where C is the constant of integration and b and c are lower and upper limits of integration.

Complete step-by-step solution -

To solve the given question, we have to find the value of the expression, which is given as,
\[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}\].
To do so we will try to integrate the given expression by using the formula of integral of \[{{e}^{-x}}\].
Formula to integrate \[{{e}^{-x}}\] or \[{{e}^{x}}\] is given by,
\[\int\limits_{{}}^{{}}{{{e}^{ax}}dx}=\dfrac{{{e}^{ax}}}{a}+C\], where a is a real number and C is the constant of integration.
Here in the above formula, the integral was used without the limits of integration.
Now we will define the integral formula of \[{{e}^{ax}}\] with the limits of integration included in it.
\[\begin{align}
  & \int\limits_{b}^{c}{{{e}^{ax}}dx}={{\left[ \dfrac{{{e}^{ax}}}{a} \right]}^{c}}_{b}+C \\
 & \Rightarrow \int\limits_{b}^{c}{{{e}^{ax}}dx}=\dfrac{1}{a}\left[ {{e}^{ac}}-{{e}^{ab}} \right]+C \\
\end{align}\]
Where b and c are limits of integration and C is the constant of integration.
Putting the value of a as -1, b as 0 and c as infinity in above formula to get the value of required integral which is given as \[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}\] we get,
\[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}={{\left[ \dfrac{{{e}^{-1x}}}{-1} \right]}^{\infty }}_{0}+C\]
Substituting the value of upper and lower limits in the above integral we get,
\[\begin{align}
  & \int\limits_{0}^{\infty }{{{e}^{-x}}dx}={{\left[ \dfrac{{{e}^{-1x}}}{-1} \right]}^{\infty }}_{0}+C \\
 & \Rightarrow \int\limits_{0}^{\infty }{{{e}^{-x}}dx}=\dfrac{1}{-1}\left[ {{e}^{-\infty }}-{{e}^{0}} \right]+C \\
\end{align}\]
Here, in above the value of constant of integration is C.
We know that the value of \[{{e}^{0}}=1\] and also that the value of \[{{e}^{-\infty }}=0\].
Substituting the value of these two expressions in the above obtained equation we get,
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{\infty }{{{e}^{-x}}dx}=\dfrac{1}{-1}\left[ {{e}^{-\infty }}-{{e}^{0}} \right]+C \\
 & \Rightarrow \int\limits_{0}^{\infty }{{{e}^{-x}}dx}=\dfrac{1}{-1}\left[ 0-1 \right]+C \\
 & \Rightarrow \int\limits_{0}^{\infty }{{{e}^{-x}}dx}=1+C \\
\end{align}\]
Therefore, we get the result as the value of \[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}\] as 1+C, where C is the constant of integration.
Generally, the constant of integration is taken as 0. So applying C = 0 in above we get the required answer as,
 \[\int\limits_{0}^{\infty }{{{e}^{-x}}dx}=1\]


Note: The possibility of error in these types of question can be at the point where you have to substitute the value of a as -1. Always remember that there will be a -1 coming in the denominator and that cannot be neglected.