What is the integral of 1/5 e^{(-y)/5} ?

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

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integral of 1/5 e^{(-y)/5}

\int \frac{1}{5} e^{\frac{- y}{5}} d y

- e^{- \frac{y}{5}} + C

Solution steps

\int \frac{1}{5} e^{\frac{- y}{5}} d y

Take the constant out:

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

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

Simplify

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - \frac{y}{5}

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

Simplify

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

= - e^{- \frac{y}{5}}

Add a constant to the solution

= - e^{- \frac{y}{5}} + C

\int \frac{ln ( y ^{2} )}{3 y} d y

y = \frac{10 x}{x ^{2} - 6}, (4, 4)

x \to 1 lim (\frac{ln ( x )}{x - 1})

\int - 20 x d x

t y^{^{'}} + 2 y + t = 0

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