What is the integral of 2^{7-x} ?

The integral of 2^{7-x} is -(2^{-x+7})/(ln(2))+C

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integral of 2^{7-x}

\int 2^{7 - x} d x

- \frac{2 ^{- x + 7}}{ln ( 2 )} + C

Solution steps

\int 2^{7 - x} d x

2^{7 - x} = 2^{7} \cdot 2^{- x}

= \int 2^{7} \cdot 2^{- x} d x

Take the constant out:

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

= 2^{7} \cdot \int 2^{- x} d x

Simplify

= 128 \cdot \int 2^{- x} d x

Apply u-substitution

: \int - 2^{u} d u

= 128 \cdot \int - 2^{u} d u

Take the constant out:

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

= 128 (- \int 2^{u} d u)

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

= 128 (- \frac{2 ^{u}}{ln ( 2 )})

Substitute back

u = - x

= 128 (- \frac{2 ^{- x}}{ln ( 2 )})

Simplify

128 (- \frac{2 ^{- x}}{ln ( 2 )}) : - \frac{2 ^{- x + 7}}{ln ( 2 )}

= - \frac{2 ^{- x + 7}}{ln ( 2 )}

Add a constant to the solution

= - \frac{2 ^{- x + 7}}{ln ( 2 )} + C

\int \frac{e ^{- x}}{cos ^{2} ( e ^{- x} )} d x

ln (x^{2} - 7)

\int \frac{x - 1}{1 - x ^{2}} d x

\frac{d}{d x} f (x) = x^{2} + 8 x + 7

x \to 0 lim (\frac{7 x ^{5} + 5 x ^{2} + 12 x}{3 x ^{5} + 4 x})

What is the integral of 2^{7-x} ?
The integral of 2^{7-x} is -(2^{-x+7})/(ln(2))+C