What is the integral of 2te^{-t^2} ?

The integral of 2te^{-t^2} is -e^{-t^2}+C

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integral of 2te^{-t^2}

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

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

Solution steps

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

Take the constant out:

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

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

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 = - t^{2}

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

Simplify

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

= - e^{- t^{2}}

Add a constant to the solution

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

\int_{0}^{1} x 1 - x^{2} d x

\int \frac{2 u - 1}{- u + 2} d u

\frac{d}{d x} (\frac{( x + 1 ) ^{2}}{x + 3})

\frac{d}{d x} (e^{\frac{1}{3} x})

e^{- x} - e^{- x} x