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

Lösung

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

x^{2} (2 x^{10} - 4 x^{7} - 2 x^{6} + 7 x^{4} - 3 x^{2} - x + 1)

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die Potenzregel an

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

Berechne die Grenzen:

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

= \frac{1}{2} (\frac{( 2 x ^{3} - 1 ) ^{4}}{4} - \frac{( 2 x ^{2} - 1 ) ^{4}}{4})

Vereinfache

= x^{2} (2 x^{10} - 4 x^{7} - 2 x^{6} + 7 x^{4} - 3 x^{2} - x + 1)

\int_{1}^{2} \frac{2}{9 + 4 x ^{2}} d x

\int_{1}^{3} 8 d x

\int_{0}^{\frac{π}{2}} cos (2 x - \frac{π}{2}) d x

\int_{0}^{1} r d r

\int_{0}^{4} 1 + 4 x^{2} d x