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

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

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integral from 0 to 1 of te^{(-t^2)/2}

\int_{0}^{1} t e^{\frac{- t ^{2}}{2}} d t

1 - \frac{1}{e}

Decimal

0.39346 \dots

Solution steps

\int_{0}^{1} t e^{\frac{- t ^{2}}{2}} d t

Simplify

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

Apply u-substitution

: \int_{0}^{- \frac{1}{2}} - e^{u} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

\int_{1}^{t a n (5 x)} 1 - t d t

\int_{0}^{2} (5 x^{2} - 6 x + 7) d x

\int_{0}^{6} \frac{185}{( x + 1 ) ^{3}} d x

\int_{0}^{4} r d r

\int_{1}^{x} e^{t} d t

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