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

Lösung

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

\frac{8}{3}

Dezimale

2.66666 \dots

Schritte zur Lösung

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

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

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

= 2 + \frac{2}{3}

Vereinfache

2 + \frac{2}{3} : \frac{8}{3}

= \frac{8}{3}

\int_{0}^{1} (- 3 x^{2} + 4 x + 3) d x

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

\int_{- 1}^{2} 8 - x^{2} - x d x

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

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