integral from x^4 to x^5 of (2t-1)^3

Lösung

\int_{x^{4}}^{x^{5}} (2 t - 1)^{3} d t

x^{4} (2 x^{16} - 2 x^{12} - 4 x^{11} + 4 x^{8} + 3 x^{6} - 3 x^{4} - x + 1)

Schritte zur Lösung

\int_{x^{4}}^{x^{5}} (2 t - 1)^{3} d t

Wende U-Substitution an

= \int_{2 x^{4} - 1}^{2 x^{5} - 1} \frac{u ^{3}}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{2 x^{4} - 1}^{2 x^{5} - 1} u^{3} d u

Wende die Potenzregel an

= \frac{1}{2} [\frac{u ^{4}}{4}]_{2 x^{4} - 1}^{2 x^{5} - 1}

Berechne die Grenzen:

\frac{( 2 x ^{5} - 1 ) ^{4}}{4} - \frac{( 2 x ^{4} - 1 ) ^{4}}{4}

= \frac{1}{2} (\frac{( 2 x ^{5} - 1 ) ^{4}}{4} - \frac{( 2 x ^{4} - 1 ) ^{4}}{4})

Vereinfache

= x^{4} (2 x^{16} - 2 x^{12} - 4 x^{11} + 4 x^{8} + 3 x^{6} - 3 x^{4} - x + 1)

\int_{- 4}^{- 3} x^{2} - (2 x + 15) d x

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

\int_{0}^{2} 3 x^{3} d x

\int_{0}^{1} 3^{x} e^{x} d x

\int_{1}^{x^{2}} e^{- t^{2}} d t