integral from 0 to 1 of (2e^{x/3+4}+5)

Lösung

\int_{0}^{1} (2 e^{\frac{x}{3} + 4} + 5) d x

6 e^{4} (3 e - 1) + 5

Dezimale

134.59823 \dots

Schritte zur Lösung

\int_{0}^{1} 2 e^{\frac{x}{3} + 4} + 5 d x

e^{\frac{x}{3} + 4} = e^{4} e^{\frac{x}{3}}

= \int_{0}^{1} 2 e^{4} e^{\frac{x}{3}} + 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_{0}^{1} 2 e^{4} e^{\frac{x}{3}} d x + \int_{0}^{1} 5 d x

\int_{0}^{1} 2 e^{4} e^{\frac{x}{3}} d x = 6 e^{4} (3 e - 1)

\int_{0}^{1} 5 d x = 5

= 6 e^{4} (3 e - 1) + 5

\int_{0}^{4} 2 x^{2} d x

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\int_{0}^{π} x sin (x^{2}) d x