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

Lösung

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

\frac{2 ( 8 - 3 3 )}{9}

Dezimale

0.62307 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{8}{3} - 3

= \frac{2}{3} (\frac{8}{3} - 3)

Vereinfache

= \frac{2 ( 8 - 3 3 )}{9}

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

\int_{0}^{π} sin (t) cos (n t) d t

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

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

\int_{- 3}^{3} - x^{2} + 9 d x