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

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

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integral from 0 to sqrt(t of)xe^{-x^2}

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

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

Solution steps

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

Apply u-substitution

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

= \int_{0}^{- t} - \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}^{- t} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{- t} - 1

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

(\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})^{^{'}}

\frac{d}{d x} (e^{\frac{x ^{2}}{2}} x)

\int e^{- 3 y} d y

x \to 6 lim (x + 6)

\int \frac{1}{3 9 x - 1} d x

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