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

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

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integral of (e^{-x})/1

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

- e^{- x} + C

Solution steps

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

Simplify

= \int e^{- x} d x

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 = - x

= - e^{- x}

Add a constant to the solution

= - e^{- x} + C

(- 8, - 3), (- 12, - 3)

\frac{\partial}{\partial x} (7 x - y)

2 y \frac{d y}{d x} + 2 = y^{2} + 2 x

\frac{d y}{d x} = \frac{x}{y}

y^{^{''}} + 8 y^{^{'}} + 20 y = 40, y (0) = 0, y^{^{'}} (0) = 0