integral from 0 to 2 of (ti+e^tj-te^tk)

Lösung

\int_{0}^{2} (t i + e^{t} j - t e^{t} k) d t

(e^{2} j - j - e^{2} k - k) + 2 i

Schritte zur Lösung

\int_{0}^{2} t i + e^{t} j - t e^{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_{0}^{2} t i d t + \int_{0}^{2} e^{t} j d t - \int_{0}^{2} t e^{t} k d t

\int_{0}^{2} t i d t = i 2

\int_{0}^{2} e^{t} j d t = j (e^{2} - 1)

\int_{0}^{2} t e^{t} k d t = k (e^{2} + 1)

= i 2 + j (e^{2} - 1) - k (e^{2} + 1)

Schreibe

i 2 + j (e^{2} - 1) - k (e^{2} + 1)

in der Standard komplexen Form um:

(e^{2} j - j - e^{2} k - k) + 2 i

= (e^{2} j - j - e^{2} k - k) + 2 i

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