integral from 0 to 7 of 5x*\sqrt[3]{x+1}

Lösung

\int_{0}^{7} 5 x \cdot 3 x + 1 d x

\frac{6045}{28}

Dezimale

215.89285 \dots

Schritte zur Lösung

\int_{0}^{7} 5 x 3 x + 1 d x

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{7} x 3 x + 1 d x

Wende U-Substitution an

= 5 \cdot \int_{1}^{8} (u - 1) 3 u d u

Multipliziere aus

(u - 1) 3 u : u^{\frac{4}{3}} - 3 u

= 5 \cdot \int_{1}^{8} u^{\frac{4}{3}} - 3 u d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int_{1}^{8} u^{\frac{4}{3}} d u - \int_{1}^{8} 3 u d u)

\int_{1}^{8} u^{\frac{4}{3}} d u = \frac{381}{7}

\int_{1}^{8} 3 u d u = \frac{45}{4}

= 5 (\frac{381}{7} - \frac{45}{4})

Vereinfache

5 (\frac{381}{7} - \frac{45}{4}) : \frac{6045}{28}

= \frac{6045}{28}

\int_{0}^{1} (6 x + x) d x

\int_{1}^{e^{π}} sin (ln (x)) d x

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

\int_{0}^{1} sin (t) \cdot cos (t) d t

\int_{- 2}^{\infty} sin (x) d x