integral from 1 to 4 of (2x+1)

Lösung

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

18

Schritte zur Lösung

\int_{1}^{4} 2 x + 1 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_{1}^{4} 2 x d x + \int_{1}^{4} 1 d x

\int_{1}^{4} 2 x d x = 15

\int_{1}^{4} 1 d x = 3

= 15 + 3

Vereinfache

= 18

x \to - 2 lim (8 + x^{2})

n = 1 \sum \infty (- \frac{1}{2})^{n}

y^{^{''}} - y^{^{'}} - 2 y = 8, y (0) = 0, y^{^{'}} (0) = 5

x \to 3 lim (4^{x^{3}} - 8 x)

x \to π - lim (sin (x))