integral from 2 to B of (4x-3)^3

Lösung

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

16 B^{4} - 48 B^{3} + 54 B^{2} - 27 B - 34

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{5}^{4 B - 3} \frac{u ^{3}}{4} d u

Entferne die Konstante:

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

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

Wende die Potenzregel an

= \frac{1}{4} [\frac{u ^{4}}{4}]_{5}^{4 B - 3}

Berechne die Grenzen:

\frac{( 4 B - 3 ) ^{4}}{4} - \frac{625}{4}

= \frac{1}{4} (\frac{( 4 B - 3 ) ^{4}}{4} - \frac{625}{4})

Vereinfache

= 16 B^{4} - 48 B^{3} + 54 B^{2} - 27 B - 34

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