What is the integral from 0 to infinity of te^{-t^2} ?

The integral from 0 to infinity of te^{-t^2} is 1/2

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integral from 0 to infinity of te^{-t^2}

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Compute the indefinite integral:

\int t e^{- t^{2}} d t = - \frac{1}{2} e^{- t^{2}} + C

Compute the boundaries:

\int_{0}^{\infty} t e^{- t^{2}} d t = 0 - (- \frac{1}{2})

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

\frac{d}{d x} (\frac{x}{x - 9})

\int (3 x - 5)^{\frac{1}{2}} d x

y^{^{''}} = 6 x - 12

y = 6 x^{4} - 6 x^{2}, y = 3 x^{2}

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 2 x = sin (2 t)

What is the integral from 0 to infinity of te^{-t^2} ?
The integral from 0 to infinity of te^{-t^2} is 1/2