What is the integral of 10^{-x} ?

The integral of 10^{-x} is -(10^{-x})/(ln(10))+C

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integral of 10^{-x}

\int 1 0^{- x} d x

- \frac{1 0 ^{- x}}{ln ( 10 )} + C

Solution steps

\int 1 0^{- x} d x

Apply u-substitution

: \int - 1 0^{u} d u

= \int - 1 0^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int 1 0^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - \frac{1 0 ^{u}}{ln ( 10 )}

Substitute back

u = - x

= - \frac{1 0 ^{- x}}{ln ( 10 )}

Add a constant to the solution

= - \frac{1 0 ^{- x}}{ln ( 10 )} + C

sin (t) cos (2 t)

8 sec (x) - 7 x

\frac{d}{d x} (x^{3} 3 x)

\int \frac{x ^{2} - 2}{( x - 2 ) ( x + 1 ) ^{3}} d x

x^{3} - 7 x

What is the integral of 10^{-x} ?
The integral of 10^{-x} is -(10^{-x})/(ln(10))+C