integral from 1 to 3 of-x^3+4x^2-3x

Lösung

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

\frac{8}{3}

Dezimale

2.66666 \dots

Schritte zur Lösung

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

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

\int_{1}^{3} 3 x d x = 12

= - 20 + \frac{104}{3} - 12

Vereinfache

- 20 + \frac{104}{3} - 12 : \frac{8}{3}

= \frac{8}{3}

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

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

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

\int_{- \infty}^{0} \frac{1}{x - 1} d x

\int_{0}^{π} sin^{6} (\frac{x}{2}) d x