How do you evaluate the integral $\int 10^{- x} d x$ $int10^(-x) dx$ ?

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Answer 1 · 2014-09-22T23:19:38

$\int 10^{- x} d x = - \frac{10^{- x}}{ln 10} + C$ $int 10^{-x}dx=-10^{-x}/ln10+C$

By rewriting to base e,

$\int 10^{- x} d x = \int e^{- (ln 10) x} d x$ $int10^{-x}dx=int e^{-(ln10)x}dx$

by the substitution $u = - (ln 10) x \Rightarrow d x = \frac{d u}{- ln 10}$ $u=-(ln10)x Rightarrow dx={du}/{-ln10}$ ,

$= - \frac{1}{ln 10} \int e^{u} d u$ $=-1/{ln10}int e^u du$

by exponential rule,

$= - \frac{1}{ln 10} e^{u} + C$ $=-1/{ln10}e^u+C$

by putting $u = - (ln 10) x$ $u=-(ln10)x$ back in,

$= - \frac{1}{ln 10} e^{- (ln 10) x} + C$ $=-1/{ln10}e^{-(ln10)x}+C$

by rewriting back to base 10,

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

How do you evaluate the integral ∫10−xdxint10^(-x) dx?