integral from 0 to 7 of xe^{-4x}

Lösung

\int_{0}^{7} x e^{- 4 x} d x

- \frac{- e ^{28} + 29}{16 e ^{28}}

Dezimale

0.06249 \dots

Schritte zur Lösung

\int_{0}^{7} x e^{- 4 x} d x

Wende U-Substitution an

= \int_{0}^{- 28} \frac{e ^{u} u}{16} d u

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

= - \int_{- 28}^{0} \frac{e ^{u} u}{16} d u

Entferne die Konstante:

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

= - \frac{1}{16} \cdot \int_{- 28}^{0} e^{u} u d u

Wende die partielle Integration an

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 28}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 28}^{0}

Berechne die Grenzen:

- 1 + \frac{29}{e ^{28}}

= - \frac{1}{16} (- 1 + \frac{29}{e ^{28}})

Vereinfache

= - \frac{- e ^{28} + 29}{16 e ^{28}}

t an g e n t f (x) = (2 x + 1)^{\frac{1}{2}}, at x = 2

\frac{d}{d x} (\frac{4 x}{1 + x ^{3}})

\frac{d}{d x} (2^{x} (π x^{2} + cos (x)))

x \to 0 lim (\frac{sin ^{2} ( 6 x )}{7})

in v erse l a pl a ce \frac{1}{s ( s ^{2} - 1 )}