${\int}_{0}^{1}{2}^{-0}d0$
Now, using the formula of integration as, $\int {a}^{x}dx=\frac{{a}^{x}}{\mathrm{log}a}+c$ to integrate the given integral.
$\int}_{0}^{1}{2}^{-0}d0=-{\left[\frac{{2}^{-0}}{\mathrm{log}\left(2\right)}\right]}_{0}^{1$ $=-{\left[\frac{1}{{2}^{0}\mathrm{log}\left(2\right)}\right]}_{0}^{1}$ $=-\frac{1}{{2}^{1}\mathrm{log}\left(2\right)}+\frac{1}{{2}^{0}\mathrm{log}\left(2\right)}$ $=-\frac{1}{2\mathrm{log}\left(2\right)}+\frac{1}{\mathrm{log}\left(2\right)}$ $=\frac{-1+2}{2\mathrm{log}\left(2\right)}$ $=\frac{1}{2\mathrm{log}\left(2\right)}$
Therefore, the value of the integral is $\frac{1}{2\mathrm{log}\left(2\right)}$.