integral of (e^ti+j+cos(t)k)

Lösung

\int (e^{t} i + j + cos (t) k) d t

i e^{t} + j t + k sin (t) + C

Schritte zur Lösung

\int e^{t} i + j + cos (t) k 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 e^{t} i d t + \int j d t + \int cos (t) k d t

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

\int j d t = j t

\int cos (t) k d t = k sin (t)

= i e^{t} + j t + k sin (t)

Füge eine Konstante zur Lösung hinzu

= i e^{t} + j t + k sin (t) + C

\frac{\partial}{\partial y} (2 x e^{x + y} + x^{2} e^{x + y} - y^{3} sin (x y))

in v erse l a pl a ce \frac{1}{( s ^{2} + 3 s )}

t an g e n t f (x) = \frac{2 x}{1 + x ^{2}}, at x = 4

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} + 2 e^{x} + e^{- x}

\int 2 + sin (x) d x