integral from 0 to 1 of x^2sqrt(1+x^3)

Lösung

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

\frac{4 2 - 2}{9}

Dezimale

0.40631 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{2} \frac{u}{3} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{1}^{2} u d u

Wende die Potenzregel an

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2}

Berechne die Grenzen:

\frac{4 2}{3} - \frac{2}{3}

= \frac{1}{3} (\frac{4 2}{3} - \frac{2}{3})

Vereinfache

= \frac{4 2 - 2}{9}

\int_{x}^{2 x} t^{3} d t

\int_{- 2}^{2} 8 d x

\int_{0}^{13} 26 x - x^{2} d x

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

\int_{0}^{2} 2 x e^{- x^{2}} d x