What is the integral of 2ye^{-y^2} ?

The integral of 2ye^{-y^2} is -e^{-y^2}+C

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integral of 2ye^{-y^2}

\int 2 y e^{- y^{2}} d y

- e^{- y^{2}} + C

Solution steps

\int 2 y e^{- y^{2}} d y

Take the constant out:

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

= 2 \cdot \int y e^{- y^{2}} d y

Apply u-substitution

: \int - \frac{e ^{u}}{2} d u

= 2 \cdot \int - \frac{e ^{u}}{2} d u

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{2} e^{u})

Substitute back

u = - y^{2}

= 2 (- \frac{1}{2} e^{- y^{2}})

Simplify

2 (- \frac{1}{2} e^{- y^{2}}) : - e^{- y^{2}}

= - e^{- y^{2}}

Add a constant to the solution

= - e^{- y^{2}} + C

\frac{1}{s ^{2}} e^{- s}

x \to 2 lim (2 x + 1)

\frac{\partial}{\partial y} (2^{x} + 2^{y})

x \to 2 lim (\frac{x - 3}{x ^{2} - 9})

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + e ^{y} + e ^{z}})