What is the integral from 0 to 1 of xe^{-5x} ?

The integral from 0 to 1 of xe^{-5x} is -(-e^5+6)/(25e^5)

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integral from 0 to 1 of xe^{-5x}

\int_{0}^{1} x e^{- 5 x} d x

- \frac{- e ^{5} + 6}{25 e ^{5}}

Decimal Notation

0.03838 \dots

Solution steps

\int_{0}^{1} x e^{- 5 x} d x

Apply u-substitution

: \int_{0}^{- 5} \frac{e ^{u} u}{25} d u

= \int_{0}^{- 5} \frac{e ^{u} u}{25} d u

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

= - \int_{- 5}^{0} \frac{e ^{u} u}{25} d u

Take the constant out:

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

= - \frac{1}{25} \cdot \int_{- 5}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 5}^{0}

= - \frac{1}{25} [e^{u} u - \int e^{u} d u]_{- 5}^{0}

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

= - \frac{1}{25} [e^{u} u - e^{u}]_{- 5}^{0}

Compute the boundaries:

- 1 + \frac{6}{e ^{5}}

= - \frac{1}{25} (- 1 + \frac{6}{e ^{5}})

Simplify

= - \frac{- e ^{5} + 6}{25 e ^{5}}

\frac{d}{d x} ((x + 4)^{3} (x - 3)^{6})

x \to 4 lim (2 - x)

y = 3 x - 7 x

x + 4 y + 2 x y

\int 4 x + 10 d x

What is the integral from 0 to 1 of xe^{-5x} ?
The integral from 0 to 1 of xe^{-5x} is -(-e^5+6)/(25e^5)