integral from 0 to t of 1/(e^x)

Lösung

\int_{0}^{t} \frac{1}{e ^{x}} d x

- \frac{1}{e ^{t}} + 1

Schritte zur Lösung

\int_{0}^{t} \frac{1}{e ^{x}} d x

Wende U-Substitution an

= \int_{1}^{e^{t}} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= [- \frac{1}{u}]_{1}^{e^{t}}

Berechne die Grenzen:

- \frac{1}{e ^{t}} + 1

= - \frac{1}{e ^{t}} + 1

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