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

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

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integral of 10^{-y}

\int 1 0^{- y} d y

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

Solution steps

\int 1 0^{- y} d y

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 = - y

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{3} - 15 x^{2} + 36 x - 10)

\int (t - \frac{1}{t}) d t

\frac{d}{d x} (sin (7 x + ln (5 x)))

x \to 0 lim (\frac{1}{x ^{2} ( x + 7 )})

t \cdot cos (k t)

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