integral from 0 to 3 of 9e^{-2x}

Lösung

\int_{0}^{3} 9 e^{- 2 x} d x

\frac{9 ( e ^{6} - 1 )}{2 e ^{6}}

Dezimale

4.48884 \dots

Schritte zur Lösung

\int_{0}^{3} 9 e^{- 2 x} d x

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{3} e^{- 2 x} d x

Wende U-Substitution an

= 9 \cdot \int_{0}^{- 6} - \frac{1}{2} e^{u} d u

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

= 9 (- \int_{- 6}^{0} - \frac{1}{2} e^{u} d u)

Entferne die Konstante:

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

= 9 (- (- \frac{1}{2} \cdot \int_{- 6}^{0} e^{u} d u))

Nutze das gemeinsame Integral :

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

= 9 (- (- \frac{1}{2} [e^{u}]_{- 6}^{0}))

Vereinfache

9 (- (- \frac{1}{2} [e^{u}]_{- 6}^{0})) : \frac{9}{2} [e^{u}]_{- 6}^{0}

= \frac{9}{2} [e^{u}]_{- 6}^{0}

Berechne die Grenzen:

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

= \frac{9}{2} (1 - \frac{1}{e ^{6}})

Vereinfache

= \frac{9 ( e ^{6} - 1 )}{2 e ^{6}}

\int_{x}^{y} 2^{l n (t)} d t

\int_{- 2}^{2} x^{2} + a x^{3} + b x^{4} d x

\int_{0.5}^{1} 1 d x

\int_{0}^{l n (2)} e^{2 x + 1} d x

\int_{6}^{5} x (x - 5)^{12} d x