integral from-1 to-4 of (2e^x+5)

Lösung

\int_{- 1}^{- 4} (2 e^{x} + 5) d x

- \frac{2}{e} + \frac{2}{e ^{4}} - 15

Dezimale

- 15.69912 \dots

Schritte zur Lösung

\int_{- 1}^{- 4} 2 e^{x} + 5 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4}^{- 1} 2 e^{x} + 5 d x

Wende die Summenregel an:

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

= - (\int_{- 4}^{- 1} 2 e^{x} d x + \int_{- 4}^{- 1} 5 d x)

\int_{- 4}^{- 1} 2 e^{x} d x = 2 (\frac{1}{e} - \frac{1}{e ^{4}})

\int_{- 4}^{- 1} 5 d x = 15

= - (2 (\frac{1}{e} - \frac{1}{e ^{4}}) + 15)

Vereinfache

- (2 (\frac{1}{e} - \frac{1}{e ^{4}}) + 15) : - \frac{2}{e} + \frac{2}{e ^{4}} - 15

= - \frac{2}{e} + \frac{2}{e ^{4}} - 15

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