integral from 1 to 5 of x^3-10x^2+29x-20

Lösung

\int_{1}^{5} x^{3} - 10 x^{2} + 29 x - 20 d x

\frac{32}{3}

Dezimale

10.66666 \dots

Schritte zur Lösung

\int_{1}^{5} x^{3} - 10 x^{2} + 29 x - 20 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}^{5} x^{3} d x - \int_{1}^{5} 10 x^{2} d x + \int_{1}^{5} 29 x d x - \int_{1}^{5} 20 d x

\int_{1}^{5} x^{3} d x = 156

\int_{1}^{5} 10 x^{2} d x = \frac{1240}{3}

\int_{1}^{5} 29 x d x = 348

\int_{1}^{5} 20 d x = 80

= 156 - \frac{1240}{3} + 348 - 80

Vereinfache

156 - \frac{1240}{3} + 348 - 80 : \frac{32}{3}

= \frac{32}{3}

\frac{\partial}{\partial y} (2 x - 1)

y^{^{'}} = y^{3}

x \to 2 lim (\frac{x ^{3} - 3 x ^{2} + 4}{3 x ^{3} - 18 x ^{2} + 36 x - 24})

\int_{4}^{\infty} 9 e^{- \frac{y}{2}} d y

\frac{\partial}{\partial x} (ln (x y^{4}))