How do you find the antiderivative \[{{e}^{-2x}}\]?

Hint: The antiderivative means integration that is opposite of derivative (differentiation). So to integrate the given function we use the chain rule, let assume \[-2x\] be a new variable \[t\] now the function is in terms of \[t\] therefore we need to find \[dx\] in terms of \[dt\] then take common outside of the integration sign and integrate the remaining part as now we can evaluate that.

Formula used:
\[\begin{align}
  & \int{{{e}^{t}}}dt={{e}^{t}} \\
 & \int{af(t)dt=a\int{f(t)dt \\
{ where\, a\, is\, constant}}}
\end{align}\].

Complete step by step solution:
We have to find \[\int{{{e}^{-2x}}}dx\]
Since we know the integration of \[{{e}^{t}}\] therefore using chain rule,
Let’s assume \[-2x\] be \[t\]
\[\Rightarrow \int{{{e}^{-2x}}}dx=\int{{{e}^{t}}dx}\]
Now we need to convert \[dx\] to \[dt\]
\[\Rightarrow -2x=t\]
Now differentiating both side
\[\begin{align}
  & \Rightarrow -2dx=dt \\
 & \Rightarrow dx=-\dfrac{dt}{2} \\
\end{align}\]
Now putting this value in the above integrating term
\[\Rightarrow \int{{{e}^{-2x}}}dx=\int{{{e}^{t}}\left( \dfrac{-dt}{2} \right)}\]
Now using another property of integration
\[\Rightarrow \int{{{e}^{-2x}}}dx=\dfrac{-1}{2}\int{{{e}^{t}}dt}\]
Since we know that,
\[\int{{{e}^{t}}}dt={{e}^{t}}\]
\[\Rightarrow \int{{{e}^{-2x}}}dx=\dfrac{-1}{2}{{e}^{t}}\]
Now again replace \[t\] with original value as \[-2x\]
\[\Rightarrow \int{{{e}^{-2x}}}dx=\dfrac{-1}{2}{{e}^{-2x}}\]

Hence the antiderivative of \[{{e}^{-2x}}\] is \[\dfrac{-1}{2}{{e}^{-2x}}\].

Note: In this type of questions in which we know the integration of the same function with some different term assume it to another variable then integrate it and note that the integrating variable \[dz\] here \[z\] must be the same variable of a function.