What is the integral of-xe^{-x^2} ?

The integral of-xe^{-x^2} is 1/2 e^{-x^2}+C

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integral of-xe^{-x^2}

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

\frac{1}{2} e^{- x^{2}} + C

Solution steps

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

Take the constant out:

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

= - \int x e^{- x^{2}} d x

Apply u-substitution

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

= - \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

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

Use the common integral:

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

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

Substitute back

u = - x^{2}

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

Simplify

= \frac{1}{2} e^{- x^{2}}

Add a constant to the solution

= \frac{1}{2} e^{- x^{2}} + C

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

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

x \to \frac{1}{3} lim (∣ 3 x - 1 ∣ - 5)

3 x^{2} y^{2} d x + (2 y^{3} x + 4 y^{3}) d y = 0

5 x^{^{''}} + 25 x^{^{'}} + 1700 x = 0