Question

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

Answer 1

`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`