What is the integral from 0 to x of 2xe^{-x^2} ?

The integral from 0 to x of 2xe^{-x^2} is -e^{-x^2}+1

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integral from 0 to x of 2xe^{-x^2}

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

- e^{- x^{2}} + 1

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{x} x e^{- x^{2}} d x

Apply u-substitution

: \int_{0}^{- x^{2}} - \frac{e ^{u}}{2} d u

= 2 \cdot \int_{0}^{- x^{2}} - \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_{0}^{- x^{2}} e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{2} [e^{u}]_{0}^{- x^{2}})

Simplify

2 (- \frac{1}{2} [e^{u}]_{0}^{- x^{2}}) : - [e^{u}]_{0}^{- x^{2}}

= - [e^{u}]_{0}^{- x^{2}}

Compute the boundaries:

e^{- x^{2}} - 1

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

Simplify

= - e^{- x^{2}} + 1

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What is the integral from 0 to x of 2xe^{-x^2} ?
The integral from 0 to x of 2xe^{-x^2} is -e^{-x^2}+1