integral from-2 to 5 of |t|

Lösung

\int_{- 2}^{5} ∣ t ∣ d t

\frac{29}{2}

Dezimale

14.5

Schritte zur Lösung

\int_{- 2}^{5} ∣ t ∣ d t

Entferne die Extremwerte

= \int_{- 2}^{0} - t d t + \int_{0}^{5} t d t

\int_{- 2}^{0} - t d t = 2

\int_{0}^{5} t d t = \frac{25}{2}

= 2 + \frac{25}{2}

Vereinfache

2 + \frac{25}{2} : \frac{29}{2}

= \frac{29}{2}

\int_{e}^{e^{4}} \frac{1}{x ( ln ( x ) ) ^{3}} d x

\int_{- 2}^{0} 2 + x^{2} d x

\int_{0}^{1} 4 x e^{- 2 x^{2}} d x

\int_{1}^{2} 8 x^{3} + \frac{6}{x ^{2}} d x

\int_{- 1}^{1} (4 x^{2} - \frac{16}{3}) d x