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

Lösung

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

\frac{217}{12}

Dezimale

18.08333 \dots

Schritte zur Lösung

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

\int_{1}^{2} 3 x^{3} d x = \frac{45}{4}

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

\int_{1}^{2} 3 x d x = \frac{9}{2}

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

= \frac{45}{4} + \frac{28}{3} - \frac{9}{2} + 2

Vereinfache

\frac{45}{4} + \frac{28}{3} - \frac{9}{2} + 2 : \frac{217}{12}

= \frac{217}{12}

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

\int_{- π}^{0} sin (x) cos (n x) d x

\int_{0}^{\frac{π}{2}} sin^{2} (z) d z

\int_{0}^{7} 14 x - x^{2} d x

\int_{0}^{6} \frac{x}{x ^{2} + 1} d x