integral from 1 to 3 of (x^2+1)

Lösung

\int_{1}^{3} (x^{2} + 1) d x

\frac{32}{3}

Dezimale

10.66666 \dots

Schritte zur Lösung

\int_{1}^{3} x^{2} + 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}^{3} x^{2} d x + \int_{1}^{3} 1 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 1 d x = 2

= \frac{26}{3} + 2

Vereinfache

\frac{26}{3} + 2 : \frac{32}{3}

= \frac{32}{3}

y^{^{''}} + 3 y^{^{'}} = 6 e^{- 3 t}, y (0) = 8, y^{^{'}} (0) = - 4

\frac{d}{d x} (e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x)

y^{2} d x - (1 - x) d y = 0

x \to 0 lim (3)

\frac{d}{d x} (- 3 x^{5} + 4 x^{- 3} - 5 x^{- 2} + 5)