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

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Answer 1 · 2014-09-22T23:19:38

$int 10^{-x}dx=-10^{-x}/ln10+C$

By rewriting to base e,

$int10^{-x}dx=int e^{-(ln10)x}dx$

by the substitution $u=-(ln10)x Rightarrow dx={du}/{-ln10}$ ,

$=-1/{ln10}int e^u du$

by exponential rule,

$=-1/{ln10}e^u+C$

by putting $u=-(ln10)x$ back in,

$=-1/{ln10}e^{-(ln10)x}+C$

by rewriting back to base 10,

$=-10^{-x}/ln10+C$

How do you evaluate the integral int10^(-x) dxint10^(-x) dx?