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

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

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integral of (e^{-y})

\int (e^{- y}) d y

- e^{- y} + C

Solution steps

\int e^{- y} d y

Apply u-substitution

: \int - e^{u} d u

= \int - e^{u} d u

Take the constant out:

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

= - \int e^{u} d u

Use the common integral:

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

= - e^{u}

Substitute back

u = - y

= - e^{- y}

Add a constant to the solution

= - e^{- y} + C

\frac{( x - 8 )}{9 x ^{2} e ^{x}}

x^{^{'}} = (1 - x) \cdot 0.03 - \frac{( 0.03 x )}{( e ^{3} - 1 )}

y^{^{'}} - \frac{2 y}{x} = - x^{2} y^{2}

\int \frac{a x}{x ^{4} + b ^{4}} d x

x \to 0 lim (\frac{- 1}{5 x})