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#