integral from 0 to x of (1+sqrt(t))

Lösung

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

x + \frac{2 x ^{\frac{3}{2}}}{3}

Schritte zur Lösung

\int_{0}^{x} 1 + t 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_{0}^{x} 1 d t + \int_{0}^{x} t d t

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

\int_{0}^{x} t d t = \frac{2 x ^{\frac{3}{2}}}{3}

= x + \frac{2 x ^{\frac{3}{2}}}{3}

\int_{1}^{\infty} \frac{9}{1 + x ^{2}} d x

\int_{0}^{2 π} 8 d t

\int_{1}^{4} x x - 1 d x

\int_{0}^{x} 1 - t^{2} d t

\int_{0}^{2} 2 x + x^{2} - x^{3} d x