integral from 4 to 9 of x(2x^2+1)^6

Lösung

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

\frac{16 3 ^{7} - 3 3 ^{7}}{28}

Dezimale

1.09182 E 14

Schritte zur Lösung

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

Wende U-Substitution an

= \int_{16}^{81} \frac{( 2 u + 1 ) ^{6}}{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_{16}^{81} (2 u + 1)^{6} d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int_{33}^{163} \frac{v ^{6}}{2} d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{33}^{163} v^{6} d v

Wende die Potenzregel an

= \frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163}

Vereinfache

\frac{1}{2} \cdot \frac{1}{2} [\frac{v ^{7}}{7}]_{33}^{163} : \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

= \frac{1}{4} [\frac{v ^{7}}{7}]_{33}^{163}

Berechne die Grenzen:

\frac{16 3 ^{7} - 3 3 ^{7}}{7}

= \frac{1}{4} \cdot \frac{16 3 ^{7} - 3 3 ^{7}}{7}

Vereinfache

= \frac{16 3 ^{7} - 3 3 ^{7}}{28}

\int_{1}^{2} \frac{1}{sin ( x ) cos ( x )} d x

\int_{100}^{150} - 0.006 x^{2} + 20 x d x

\int_{0}^{2} \frac{1}{x ^{2} + 3 x + 4} d x

\int_{2}^{5} \frac{1}{( x - 2 ) ^{n}} d x

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