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

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

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

\int \frac{d}{1} (e^{- x} y) d x

- d y e^{- x} + C

Solution steps

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

Take the constant out:

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

= d y \cdot \int e^{- x} d x

Apply u-substitution

: \int - e^{u} d u

= d y \cdot \int - e^{u} d u

Take the constant out:

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

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

Use the common integral:

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

= d y (- e^{u})

Substitute back

u = - x

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

Simplify

= - d y e^{- x}

Add a constant to the solution

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

x \to 0 lim (\frac{e ^{2 x} - 1}{x})

\frac{d y}{d x} = \frac{x + y ^{2}}{2 y}, y (0) = 1

\int \frac{x ^{5}}{x ^{6} - 7} d x

f (x) = arccos (4 x^{2})

θ \to 0 lim (\frac{1}{11})

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