integral from 0 to 1 of 16(2+4x)^3

Lösung

\int_{0}^{1} 16 (2 + 4 x)^{3} d x

1280

Schritte zur Lösung

\int_{0}^{1} 16 (2 + 4 x)^{3} d x

Entferne die Konstante:

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

= 16 \cdot \int_{0}^{1} (2 + 4 x)^{3} d x

Wende U-Substitution an

= 16 \cdot \int_{2}^{6} \frac{u ^{3}}{4} d u

Entferne die Konstante:

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

= 16 \cdot \frac{1}{4} \cdot \int_{2}^{6} u^{3} d u

Wende die Potenzregel an

= 16 \cdot \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{6}

Vereinfache

16 \cdot \frac{1}{4} [\frac{u ^{4}}{4}]_{2}^{6} : 4 [\frac{u ^{4}}{4}]_{2}^{6}

= 4 [\frac{u ^{4}}{4}]_{2}^{6}

Berechne die Grenzen:

320

= 4 \cdot 320

Vereinfache

= 1280

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

\int_{0}^{\frac{π}{4}} 6 sec (x) tan (x) d x

\int_{1}^{2} (4 x^{2} + \frac{8}{x ^{2}}) d x

\int_{- 2}^{1} (4 x^{2} + 1) d x

\int_{0}^{1} (t - 1) d t