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

Lösung

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

14 7 e - 9

Dezimale

7.14990 \dots

Schritte zur Lösung

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

Vereinfache

= \int_{0}^{1} 2 e^{\frac{x}{7}} + 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^{\frac{x}{7}} d x + \int_{0}^{1} 5 d x

\int_{0}^{1} 2 e^{\frac{x}{7}} d x = 14 (7 e - 1)

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

= 14 (7 e - 1) + 5

Vereinfache

14 (7 e - 1) + 5 : 14 7 e - 9

= 14 7 e - 9

\int_{1}^{3} 5^{3 x} d x

\int_{0}^{\frac{π}{4}} 3 sec^{2} (t) d t

\int_{0}^{3} 3 8 x + 1 d x

\int_{4}^{6} x^{2} d x

\int_{0}^{4} x ln (5 - x) d x