What is the integral from 0 to 1 of xe^{-3x} ?

The integral from 0 to 1 of xe^{-3x} is -(-e^3+4)/(9e^3)

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integral from 0 to 1 of xe^{-3x}

\int_{0}^{1} x e^{- 3 x} d x

- \frac{- e ^{3} + 4}{9 e ^{3}}

Decimal

0.08898 \dots

Solution steps

\int_{0}^{1} x e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 3} \frac{e ^{u} u}{9} d u

= \int_{0}^{- 3} \frac{e ^{u} u}{9} d u

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

= - \int_{- 3}^{0} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= - \frac{1}{9} \cdot \int_{- 3}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 3}^{0}

= - \frac{1}{9} [e^{u} u - \int e^{u} d u]_{- 3}^{0}

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

= - \frac{1}{9} [e^{u} u - e^{u}]_{- 3}^{0}

Compute the boundaries:

- 1 + \frac{4}{e ^{3}}

= - \frac{1}{9} (- 1 + \frac{4}{e ^{3}})

Simplify

= - \frac{- e ^{3} + 4}{9 e ^{3}}

\int \frac{5 t}{1 - 25 t ^{4}} d t

cot (csc (x))

g (x) = \frac{1}{x + 4}

\int_{1}^{5} \frac{1}{x - 1} d x

\frac{\partial}{\partial x} (e^{x \cdot e^{y}})

What is the integral from 0 to 1 of xe^{-3x} ?
The integral from 0 to 1 of xe^{-3x} is -(-e^3+4)/(9e^3)