integral from 0 to 5 of xe^{-0.4x}

Lösung

\int_{0}^{5} x e^{- 0.4 x} d x

3.71246 \dots

Schritte zur Lösung

\int_{0}^{5} x e^{- 0.4 x} d x

Wende U-Substitution an

= \int_{0}^{- 2} \frac{5 e ^{u} u}{0.8} d u

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

= - \int_{- 2}^{0} \frac{5 e ^{u} u}{0.8} d u

Entferne die Konstante:

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

= - \frac{5}{0.8} \cdot \int_{- 2}^{0} e^{u} u d u

Vereinfache

= - 6.25 \cdot \int_{- 2}^{0} e^{u} u d u

Wende die partielle Integration an

= - 6.25 [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

= - 6.25 [e^{u} u - e^{u}]_{- 2}^{0}

Berechne die Grenzen:

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

= - 6.25 (- 1 + \frac{3}{e ^{2}})

Vereinfache

= 3.71246 \dots

\int_{0}^{1} x^{20} d x

\int_{0}^{1} \frac{1}{( 4 x - 1 ) ^{3}} d x

\int_{3}^{6} (x + 1) ln (2 x + 4) d x

\int_{0}^{5} e^{0.25 x} d x

\int_{0}^{3} (4 x f (x^{2}) + f (3 x)) d x