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# System prepended metadata

title: PID-704

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The given integral is
$$ I=\int 2^{-x}dx \tag1$$

We solve the above integral by using the following substitution:
\begin{align*}
        &2^{-x} = t\\
        \implies -2^{-x}\cdot&\ln(2)\cdot dx=dt\\
        \implies 2^{-x}&\cdot dx =-\frac{dt}{\ln(2)}
\end{align*}

Substituting the above value in the integral (1), we have:
\begin{align*}
        I&=\int -\frac{dt}{\ln(2)}\\
        &= -\frac{1}{\ln(2)}\int dt\\
        &=-\frac{1}{\ln(2)}t +c\\
        &= \frac{-2^{-x}}{\ln(2)} +c
\end{align*}
Thus, we have
$$\int 2^{-x}dx = -\frac{2^{-x}}{\ln(2)} +c$$