# How do you evaluate the integral int10^(-x) dx?

Sep 22, 2014

$\int {10}^{- x} \mathrm{dx} = - {10}^{- x} / \ln 10 + C$

By rewriting to base e,

$\int {10}^{- x} \mathrm{dx} = \int {e}^{- \left(\ln 10\right) x} \mathrm{dx}$

by the substitution $u = - \left(\ln 10\right) x R i g h t a r r o w \mathrm{dx} = \frac{\mathrm{du}}{- \ln 10}$,

$= - \frac{1}{\ln 10} \int {e}^{u} \mathrm{du}$

by exponential rule,

$= - \frac{1}{\ln 10} {e}^{u} + C$

by putting $u = - \left(\ln 10\right) x$ back in,

$= - \frac{1}{\ln 10} {e}^{- \left(\ln 10\right) x} + C$

by rewriting back to base 10,

$= - {10}^{- x} / \ln 10 + C$