What is the integral from 0 to 1 of xe^{-8x^2} ?

The integral from 0 to 1 of xe^{-8x^2} is (e^8-1)/(16e^8)

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integral from 0 to 1 of xe^{-8x^2}

\int_{0}^{1} x e^{- 8 x^{2}} d x

\frac{e ^{8} - 1}{16 e ^{8}}

Decimal Notation

0.06247 \dots

Solution steps

\int_{0}^{1} x e^{- 8 x^{2}} d x

Apply u-substitution

: \int_{0}^{- 8} - \frac{e ^{u}}{16} d u

= \int_{0}^{- 8} - \frac{e ^{u}}{16} d u

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

= - \int_{- 8}^{0} - \frac{e ^{u}}{16} d u

Take the constant out:

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

= - (- \frac{1}{16} \cdot \int_{- 8}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{16} [e^{u}]_{- 8}^{0})

Simplify

= \frac{1}{16} [e^{u}]_{- 8}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{8}}

= \frac{1}{16} (1 - \frac{1}{e ^{8}})

Simplify

= \frac{e ^{8} - 1}{16 e ^{8}}

\frac{d y}{d x} + 9 y = 3

\int \frac{14}{1 - cos ( 7 x )} d x

y^{^{''}} + 2 y = cos (t) - 2 \cdot sin (2 t)

\int 3 e^{- 3 x} d x

y^{^{''}} - 2 y^{^{'}} + y = t^{- 3} e^{t}

What is the integral from 0 to 1 of xe^{-8x^2} ?
The integral from 0 to 1 of xe^{-8x^2} is (e^8-1)/(16e^8)