integral from 1 to x of (t+1)

Lösung

\int_{1}^{x} (t + 1) d t

\frac{x ^{2} - 1}{2} + x - 1

Schritte zur Lösung

\int_{1}^{x} t + 1 d t

Wende die Summenregel an:

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

= \int_{1}^{x} t d t + \int_{1}^{x} 1 d t

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

\int_{1}^{x} 1 d t = x - 1

= \frac{x ^{2} - 1}{2} + x - 1

\int_{0}^{1} 2 (x + y) d x

\int_{2}^{4} \frac{x}{( x + 1 ) ( x + 2 )} d x

\int_{- \infty}^{3} \frac{1}{x ^{2} - 1} d x

\int_{0}^{32} \frac{1}{5 x} d x

\int_{0}^{\infty} sin^{2} (a) d a