What is the integral from 0 to 2 of 4xe^{-2x} ?

The integral from 0 to 2 of 4xe^{-2x} is 1-5/(e^4)

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integral from 0 to 2 of 4xe^{-2x}

\int_{0}^{2} 4 x e^{- 2 x} d x

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

Decimal

0.90842 \dots

Solution steps

\int_{0}^{2} 4 x e^{- 2 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{2} x e^{- 2 x} d x

Apply u-substitution

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

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

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

= 4 (- \int_{- 4}^{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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

= - [e^{u} u - e^{u}]_{- 4}^{0}

Compute the boundaries:

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

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

Simplify

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

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\int_{1}^{4} 7 x d x

\int_{0}^{1} 5 (1 - x)^{4} d x

What is the integral from 0 to 2 of 4xe^{-2x} ?
The integral from 0 to 2 of 4xe^{-2x} is 1-5/(e^4)