integral from-1 to 0 of sqrt(3x+4)

Lösung

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

\frac{14}{9}

Dezimale

1.55555 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{4} \frac{u}{3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{1}^{4} u d u

Wende die Potenzregel an

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Berechne die Grenzen:

\frac{14}{3}

= \frac{1}{3} \cdot \frac{14}{3}

Vereinfache

= \frac{14}{9}

\int_{0}^{2} (6 t^{2} i - 4 t j + 3 k) d t

\int_{5}^{10} \frac{2 x}{x ^{2} + 1} d x

\int_{0}^{3} t 1 + 16 t^{2} d t

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

\int_{0}^{2} 3 t 4 - t^{2} d t