integral of (200e^{0.1t}+150e^{-0.3t})

Lösung

\int (200 e^{0.1 t} + 150 e^{- 0.3 t}) d t

2000 e^{0.1 t} - 499.99999 \dots e^{- 0.3 t} + C

Schritte zur Lösung

\int 200 e^{0.1 t} + 150 e^{- 0.3 t} d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 200 e^{0.1 t} d t + \int 150 e^{- 0.3 t} d t

\int 200 e^{0.1 t} d t = 2000 e^{0.1 t}

\int 150 e^{- 0.3 t} d t = - 499.99999 \dots e^{- 0.3 t}

= 2000 e^{0.1 t} - 499.99999 \dots e^{- 0.3 t}

Füge eine Konstante zur Lösung hinzu

= 2000 e^{0.1 t} - 499.99999 \dots e^{- 0.3 t} + C

\int \frac{z}{4 a ^{2} - z ^{2}} d z

\int \frac{1}{y \cdot ( 2 + y ^{3} )} d y

\int 15 tan^{2} (x) sec^{4} (x) d x

\int y 3 + y^{2} d y

\int \frac{cos ( e ^{x} )}{sin ( e ^{x} )} e^{x} d x