integral from 0 to 1 of e^{-8x}

Lösung

\int_{0}^{1} e^{- 8 x} d x

\frac{e ^{8} - 1}{8 e ^{8}}

Dezimale

0.12495 \dots

Schritte zur Lösung

\int_{0}^{1} e^{- 8 x} d x

Wende U-Substitution an

= \int_{0}^{- 8} - \frac{1}{8} e^{u} d u

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

= - \int_{- 8}^{0} - \frac{1}{8} e^{u} d u

Entferne die Konstante:

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

= - (- \frac{1}{8} \cdot \int_{- 8}^{0} e^{u} d u)

Nutze das gemeinsame Integral :

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

= - (- \frac{1}{8} [e^{u}]_{- 8}^{0})

Vereinfache

= \frac{1}{8} [e^{u}]_{- 8}^{0}

Berechne die Grenzen:

1 - \frac{1}{e ^{8}}

= \frac{1}{8} (1 - \frac{1}{e ^{8}})

Vereinfache

= \frac{e ^{8} - 1}{8 e ^{8}}

x \to \infty lim \frac{( e ^{\frac{15}{x}} - 1 )}{\frac{15}{x}}

in v erse l a pl a ce (s + \frac{3}{2})^{2}

\int e^{\frac{- x}{k}} \cdot \frac{- x}{k} d x

x \to \infty lim (x^{3} - 2 x)

d er i v a t i v e f (x) = \frac{2 x + 5}{x + 2} \cdot 2 x - 5