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

Lösung

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

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

Dezimale

0.90842 \dots

Schritte zur Lösung

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

Entferne die Konstante:

\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

Wende U-Substitution an

= 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)

Entferne die Konstante:

\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)

Wende die partielle Integration an

= 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})

Vereinfache

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}

Berechne die Grenzen:

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

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

Vereinfache

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

\int_{0}^{1} (\frac{5}{7 + 7 x ^{2}}) d x

\int_{0}^{2} x x d x

\int_{1}^{2} \frac{1 + y}{1 + y ^{2}} d y

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

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