What is the integral of 2^{9-x}-64 ?

The integral of 2^{9-x}-64 is -(2^{-x+9})/(ln(2))-64x+C

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integral of 2^{9-x}-64

\int 2^{9 - x} - 64 d x

- \frac{2 ^{- x + 9}}{ln ( 2 )} - 64 x + C

Solution steps

\int 2^{9 - x} - 64 d x

Simplify

2^{9 - x} : 2^{9} \cdot 2^{- x}

= \int 2^{9} \cdot 2^{- x} - 64 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2^{9} \cdot 2^{- x} d x - \int 64 d x

\int 2^{9} \cdot 2^{- x} d x = - \frac{2 ^{- x + 9}}{ln ( 2 )}

\int 64 d x = 64 x

= - \frac{2 ^{- x + 9}}{ln ( 2 )} - 64 x

Add a constant to the solution

= - \frac{2 ^{- x + 9}}{ln ( 2 )} - 64 x + C

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What is the integral of 2^{9-x}-64 ?
The integral of 2^{9-x}-64 is -(2^{-x+9})/(ln(2))-64x+C