integral from 1 to 2 of 2x^2sqrt(x^3+6)

Lösung

\int_{1}^{2} 2 x^{2} x^{3} + 6 d x

\frac{2 ( 28 14 - 14 7 )}{9}

Dezimale

15.05019 \dots

Schritte zur Lösung

\int_{1}^{2} 2 x^{2} x^{3} + 6 d x

Entferne die Konstante:

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

= 2 \cdot \int_{1}^{2} x^{2} x^{3} + 6 d x

Wende U-Substitution an

= 2 \cdot \int_{7}^{14} \frac{u}{3} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{3} \cdot \int_{7}^{14} u d u

Wende die Potenzregel an

= 2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

Vereinfache

2 \cdot \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14} : \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

= \frac{2}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{14}

Berechne die Grenzen:

\frac{28 14 - 14 7}{3}

= \frac{2}{3} \cdot \frac{28 14 - 14 7}{3}

Vereinfache

= \frac{2 ( 28 14 - 14 7 )}{9}

\int_{5}^{x} (\frac{1}{5} t^{2} - 1)^{7} d t

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

\int_{- 1}^{2} t d t

\int_{0}^{3} r 25 - r^{2} d r

\int_{- 1}^{1} x x^{8} + 3 d x