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

Lösung

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

\frac{125}{6}

Dezimale

20.83333 \dots

Schritte zur Lösung

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

Vereinfache

= \int_{- 1}^{4} - x^{2} + 3 x + 4 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} 3 x d x + \int_{- 1}^{4} 4 d x

\int_{- 1}^{4} x^{2} d x = \frac{65}{3}

\int_{- 1}^{4} 3 x d x = \frac{45}{2}

\int_{- 1}^{4} 4 d x = 20

= - \frac{65}{3} + \frac{45}{2} + 20

Vereinfache

- \frac{65}{3} + \frac{45}{2} + 20 : \frac{125}{6}

= \frac{125}{6}

\int_{0}^{1} x^{\frac{5}{6}} d x

\int_{0}^{1} (x + 2 1 - x^{2}) d x

\int_{0}^{1} \frac{1}{4 + 5 x - 1} d x

\int_{0}^{5} \frac{x}{x ^{2} + 1} d x

\int_{4}^{5} 1