integral from 1 to 2 of (6x^3-4x^2+2x)

Lösung

\int_{1}^{2} (6 x^{3} - 4 x^{2} + 2 x) d x

\frac{97}{6}

Dezimale

16.16666 \dots

Schritte zur Lösung

\int_{1}^{2} 6 x^{3} - 4 x^{2} + 2 x d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} 6 x^{3} d x - \int_{1}^{2} 4 x^{2} d x + \int_{1}^{2} 2 x d x

\int_{1}^{2} 6 x^{3} d x = \frac{45}{2}

\int_{1}^{2} 4 x^{2} d x = \frac{28}{3}

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

= \frac{45}{2} - \frac{28}{3} + 3

Vereinfache

\frac{45}{2} - \frac{28}{3} + 3 : \frac{97}{6}

= \frac{97}{6}

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

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

\int_{0}^{\frac{π}{6}} 1 + (tan (x))^{2} d x

\int_{1}^{2} \frac{sinh ( ln ( x ) )}{x} d x

\int_{1}^{2} 2 - y d y