What is the integral from 0 to 1 of 2te^{-2t} ?

The integral from 0 to 1 of 2te^{-2t} is -(-e^2+3)/(2e^2)

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integral from 0 to 1 of 2te^{-2t}

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

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

Decimal Notation

0.29699 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{1} t e^{- 2 t} d t

Apply u-substitution

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

= 2 \cdot \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

= 2 (- \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

= 2 (- \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}

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

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

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

Simplify

2 (- \frac{1}{4} [e^{u} u - e^{u}]_{- 2}^{0}) : - \frac{1}{2} [e^{u} u - e^{u}]_{- 2}^{0}

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

Compute the boundaries:

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

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

Simplify

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

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