integral from 1 to 5 of (x+1)^2

Lösung

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

\frac{208}{3}

Dezimale

69.33333 \dots

Schritte zur Lösung

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

Multipliziere aus

(x + 1)^{2} : x^{2} + 2 x + 1

= \int_{1}^{5} x^{2} + 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}^{5} x^{2} d x + \int_{1}^{5} 2 x d x + \int_{1}^{5} 1 d x

\int_{1}^{5} x^{2} d x = \frac{124}{3}

\int_{1}^{5} 2 x d x = 24

\int_{1}^{5} 1 d x = 4

= \frac{124}{3} + 24 + 4

Vereinfache

\frac{124}{3} + 24 + 4 : \frac{208}{3}

= \frac{208}{3}

d er i v a t i v e (t^{2} + 1) (t - \frac{1}{t})

\frac{\partial}{\partial x} (2 x^{y})

\frac{d x}{d t} = x - 11

\int_{0}^{\frac{π}{2}} x sin (2 x) d x

s l o p e (3, 4), (- 2, - 4)