What is the integral from 0 to 1 of xe^{-9x^2} ?

The integral from 0 to 1 of xe^{-9x^2} is (e^9-1)/(18e^9)

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integral from 0 to 1 of xe^{-9x^2}

\int_{0}^{1} x e^{- 9 x^{2}} d x

\frac{e ^{9} - 1}{18 e ^{9}}

Decimal

0.05554 \dots

Solution steps

\int_{0}^{1} x e^{- 9 x^{2}} d x

Apply u-substitution

= \int_{0}^{- 9} - \frac{e ^{u}}{18} d u

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

= - \int_{- 9}^{0} - \frac{e ^{u}}{18} d u

Take the constant out:

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

= - (- \frac{1}{18} \cdot \int_{- 9}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{18} [e^{u}]_{- 9}^{0})

Simplify

= \frac{1}{18} [e^{u}]_{- 9}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{9}}

= \frac{1}{18} (1 - \frac{1}{e ^{9}})

Simplify

= \frac{e ^{9} - 1}{18 e ^{9}}

What is the integral from 0 to 1 of xe^{-9x^2} ?
The integral from 0 to 1 of xe^{-9x^2} is (e^9-1)/(18e^9)