integral from 0 to 1 of 1/2 e^x+3/4 2^x

Lösung

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

\frac{e - 1}{2} + \frac{3}{4 ln ( 2 )}

Dezimale

1.94116 \dots

Schritte zur Lösung

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

\int_{0}^{1} \frac{1}{2} e^{x} d x = \frac{e - 1}{2}

\int_{0}^{1} \frac{3}{4} \cdot 2^{x} d x = \frac{3}{4 ln ( 2 )}

= \frac{e - 1}{2} + \frac{3}{4 ln ( 2 )}

\int_{- 3}^{7} (∣ x ∣) d x

\int_{0}^{\frac{π}{2}} sin (8 x) d x

\int_{0}^{π} 2 cos^{2} (t) d t

\int_{- 1}^{2} (2 - 2 ∣ x ∣) d x

\int_{0}^{1.5} (- 4 x + 6)^{\frac{3}{2}} d x