integral from 0 to 1 of 2^x+x

Lösung

\int_{0}^{1} 2^{x} + x d x

\frac{1}{ln ( 2 )} + \frac{1}{2}

Dezimale

1.94269 \dots

Schritte zur Lösung

\int_{0}^{1} 2^{x} + 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} 2^{x} d x + \int_{0}^{1} x d x

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

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

= \frac{1}{ln ( 2 )} + \frac{1}{2}

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

\int_{- 2}^{2} (x^{2}) d x

\int_{0}^{4} (x^{2} - 2 y^{2}) d x

\int_{1}^{5} (2 x + \frac{1}{x}) d x

\int_{0}^{1} e^{x} - e d x