integral from 0 to 3 of te^{-t}

Lösung

\int_{0}^{3} t e^{- t} d t

- \frac{4}{e ^{3}} + 1

Dezimale

0.80085 \dots

Schritte zur Lösung

\int_{0}^{3} t e^{- t} d t

Wende die partielle Integration an

= [- e^{- t} t - \int - e^{- t} d t]_{0}^{3}

\int - e^{- t} d t = e^{- t}

= [- e^{- t} t - e^{- t}]_{0}^{3}

Berechne die Grenzen:

- \frac{4}{e ^{3}} + 1

= - \frac{4}{e ^{3}} + 1

a re a y^{2} = 4 x, 4 x - 3 y = 4

\int_{1}^{5} \frac{x}{2 x - 1} d x

x \to 0 lim (cos (x) + sin (x))

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{x ^{2} - 1}, y (7) = 7

y^{^{'}} = 8 x^{3} + 2 \frac{y}{x}