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

Lösung

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

\frac{14}{27}

Dezimale

0.51851 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{4} \frac{u}{9} d u

Entferne die Konstante:

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

= \frac{1}{9} \cdot \int_{1}^{4} u d u

Wende die Potenzregel an

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Berechne die Grenzen:

\frac{14}{3}

= \frac{1}{9} \cdot \frac{14}{3}

Vereinfache

= \frac{14}{27}

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

\int_{4}^{3} x^{2} d x

\int_{0}^{π} (10 + x) d x

\int_{4}^{3 x + 4} \frac{t}{1 + t ^{3}} d t

\int_{- 1}^{0} x^{\frac{1}{3}} d x