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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

= - \frac{1}{2} \cdot \int_{0}^{- x^{2}} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

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

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

\int_{1}^{3} x^{2} + 4 x d x

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

\int_{16}^{20} 0.005 (20 - x) d x

\int_{0}^{x^{3}} e^{- t + 1} d t

\int_{- 1}^{1} 1 - ∣ x ∣ x d x

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