integral from-1 to 0 of |4x+3|

Lösung

\int_{- 1}^{0} ∣ 4 x + 3 ∣ d x

\frac{5}{4}

Dezimale

1.25

Schritte zur Lösung

\int_{- 1}^{0} ∣ 4 x + 3 ∣ d x

Entferne die Extremwerte

= \int_{- 1}^{- \frac{3}{4}} - 4 x - 3 d x + \int_{- \frac{3}{4}}^{0} 4 x + 3 d x

\int_{- 1}^{- \frac{3}{4}} - 4 x - 3 d x = \frac{1}{8}

\int_{- \frac{3}{4}}^{0} 4 x + 3 d x = \frac{9}{8}

= \frac{1}{8} + \frac{9}{8}

Vereinfache

\frac{1}{8} + \frac{9}{8} : \frac{5}{4}

= \frac{5}{4}

\int_{1}^{2.8} ln (x^{2}) d x

\int_{0}^{2} (15 - t) t d t

\int_{0}^{2} x \cdot arctan (\frac{x}{2}) d x

\int_{0}^{\frac{1}{4}} 4 1 - 16 x^{2} d x

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