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

Lösung

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

\frac{33}{2}

Dezimale

16.5

Schritte zur Lösung

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

\int_{1}^{4} x^{2} d x = 21

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

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

= 21 - \frac{15}{2} + 3

Vereinfache

21 - \frac{15}{2} + 3 : \frac{33}{2}

= \frac{33}{2}

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

\int_{0}^{4} \frac{2}{x + 1} d x

\int_{- 3}^{3} x^{3} + 4 x + 1 d x

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

\int_{- 2}^{x} 1 - x^{4} d x