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

The integral from 0 to 1 of xe^{-2x} is -(-e^2+3)/(4e^2)

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

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

- \frac{- e ^{2} + 3}{4 e ^{2}}

Decimal Notation

0.14849 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 2} \frac{e ^{u} u}{4} d u

= \int_{0}^{- 2} \frac{e ^{u} u}{4} d u

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

= - \int_{- 2}^{0} \frac{e ^{u} u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int_{- 2}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 2}^{0}

= - \frac{1}{4} [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

= - \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0}

Compute the boundaries:

- 1 + \frac{3}{e ^{2}}

= - \frac{1}{4} (- 1 + \frac{3}{e ^{2}})

Simplify

= - \frac{- e ^{2} + 3}{4 e ^{2}}

y = 4 x^{2} - 1, y = 0, x = \frac{- 1}{2}, x = 1

\int \frac{x + 25}{x ^{2} + 24 x + 148} d x

y^{^{'}} = \frac{( cos ( 18 x ) )}{( e ^{(18 \cdot y)} )}

\int \frac{1}{2} e^{- \frac{1}{2} x} d x

\int \frac{1}{x} + x^{2} - 2 d x

What is the integral from 0 to 1 of xe^{-2x} ?
The integral from 0 to 1 of xe^{-2x} is -(-e^2+3)/(4e^2)