integral from-1 to 1 of \sqrt[3]{t-2}

Lösung

\int_{- 1}^{1} 3 t - 2 d t

\frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

Schritte zur Lösung

\int_{- 1}^{1} 3 t - 2 d t

Wende U-Substitution an

= \int_{- 3}^{- 1} 3 u d u

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

= \frac{- 3 3 - 1 + 9 3 - 1 3 3}{4}

\int_{0}^{2} y e^{x - y} d x

\int_{0}^{5} (x^{3} - x) d x

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

\int_{0}^{1} 1 - x^{2} \cdot x d x

\int_{1}^{5} e^{- x} d x