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

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

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integral of e^{-(x+y)}

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

- e^{- x - y} + C

Solution steps

\int e^{- (x + 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 = - (x + y)

= - e^{- (x + y)}

Simplify

= - e^{- x - y}

Add a constant to the solution

= - e^{- x - y} + C

\frac{d}{d x} (\frac{2}{2 - cos ( π x )})

\frac{\partial}{\partial x} (- x ln (x) + x ln (y) + y)

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

\frac{d}{d x} (\frac{- x ^{2} - 2 x + 1}{( x ^{2} + 1 ) ^{2}})

x \to 2 - lim (x^{2} - x + 2)