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

Lösung

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

\frac{26}{3}

Dezimale

8.66666 \dots

Schritte zur Lösung

\int_{1}^{3} - x^{3} + 4 x^{2} - 3 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 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 d x = 6

= - 20 + \frac{104}{3} - 6

Vereinfache

- 20 + \frac{104}{3} - 6 : \frac{26}{3}

= \frac{26}{3}

\int_{7}^{x} sin (t^{4}) d t

\int_{0}^{3} \frac{x ^{3}}{9} d x

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

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

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