What is the integral of (e^{4-x})^5 ?

The integral of (e^{4-x})^5 is -1/5 e^{20-5x}+C

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integral of (e^{4-x})^5

\int (e^{4 - x})^{5} d x

- \frac{1}{5} e^{20 - 5 x} + C

Solution steps

\int (e^{4 - x})^{5} d x

(e^{4 - x})^{5} = (e^{4})^{5} (e^{- x})^{5}

= \int (e^{4})^{5} (e^{- x})^{5} d x

Take the constant out:

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

= (e^{4})^{5} \cdot \int (e^{- x})^{5} d x

(e^{4})^{5} = e^{20}

= e^{20} \cdot \int (e^{- x})^{5} d x

Simplify

(e^{- x})^{5} : e^{- 5 x}

= e^{20} \cdot \int e^{- 5 x} d x

Apply u-substitution

: \int - \frac{1}{5} e^{u} d u

= e^{20} \cdot \int - \frac{1}{5} e^{u} d u

Take the constant out:

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

= e^{20} (- \frac{1}{5} \cdot \int e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= e^{20} (- \frac{1}{5} e^{u})

Substitute back

u = - 5 x

= e^{20} (- \frac{1}{5} e^{- 5 x})

Simplify

e^{20} (- \frac{1}{5} e^{- 5 x}) : - \frac{1}{5} e^{20 - 5 x}

= - \frac{1}{5} e^{20 - 5 x}

Add a constant to the solution

= - \frac{1}{5} e^{20 - 5 x} + C

\int_{- 3}^{3} - 2 x^{2} + 18 d x

\frac{d}{d x} (100 e^{x})

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

y^{^{''}} + y = tan (x)

f (x) = x^{2} + x, at x = 3

What is the integral of (e^{4-x})^5 ?
The integral of (e^{4-x})^5 is -1/5 e^{20-5x}+C