integral from 0 to 10 of 20te^{-0.6t}

Lösung

\int_{0}^{10} 20 t e^{- 0.6 t} d t

54.59159 \dots

Schritte zur Lösung

\int_{0}^{10} 20 t e^{- 0.6 t} d t

Entferne die Konstante:

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

= 20 \cdot \int_{0}^{10} t e^{- 0.6 t} d t

Wende U-Substitution an

= 20 \cdot \int_{0}^{- 6} \frac{5 e ^{u} u}{1.8} d u

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

= 20 (- \int_{- 6}^{0} \frac{5 e ^{u} u}{1.8} d u)

Entferne die Konstante:

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

= 20 (- \frac{5}{1.8} \cdot \int_{- 6}^{0} e^{u} u d u)

Vereinfache

= 20 (- 2.77777 \dots \cdot \int_{- 6}^{0} e^{u} u d u)

Wende die partielle Integration an

= 20 (- 2.77777 \dots [e^{u} u - \int e^{u} d u]_{- 6}^{0})

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

= 20 (- 2.77777 \dots [e^{u} u - e^{u}]_{- 6}^{0})

Vereinfache

= - 55.55555 \dots [e^{u} u - e^{u}]_{- 6}^{0}

Berechne die Grenzen:

- 1 + \frac{7}{e ^{6}}

= - 55.55555 \dots (- 1 + \frac{7}{e ^{6}})

Vereinfache

= 54.59159 \dots

\frac{d}{d x} (arccos (\frac{x}{2}))

y^{^{'}} = \frac{( x ^{2} + 3 y ^{2} )}{2 x y}

x \to 3 lim (x - 13)

y^{^{''}} - y^{^{'}} + y = 13 sin (t)

\frac{d}{d x} (sin (2 x^{3} - 3))