What is the integral from 0 to 1 of xe^{-4x} ?

The integral from 0 to 1 of xe^{-4x} is -(-e^4+5)/(16e^4)

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integral from 0 to 1 of xe^{-4x}

\int_{0}^{1} x e^{- 4 x} d x

- \frac{- e ^{4} + 5}{16 e ^{4}}

Decimal Notation

0.05677 \dots

Solution steps

\int_{0}^{1} x e^{- 4 x} d x

Apply u-substitution

: \int_{0}^{- 4} \frac{e ^{u} u}{16} d u

= \int_{0}^{- 4} \frac{e ^{u} u}{16} d u

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

= - \int_{- 4}^{0} \frac{e ^{u} 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_{- 4}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 4}^{0}

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 4}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 4}^{0}

Compute the boundaries:

- 1 + \frac{5}{e ^{4}}

= - \frac{1}{16} (- 1 + \frac{5}{e ^{4}})

Simplify

= - \frac{- e ^{4} + 5}{16 e ^{4}}

\int \frac{1}{x 25 x ^{2} + 16} d x

x + y \frac{d y}{d x} = 0

\frac{d}{d x} (3 e^{2 x} + 2 e^{2 x} x)

f (x) = x^{2} + 7, 16, at x = 3

\int \frac{e ^{- 3 t}}{t} d t

What is the integral from 0 to 1 of xe^{-4x} ?
The integral from 0 to 1 of xe^{-4x} is -(-e^4+5)/(16e^4)