integral from 0 to 7 of 4x\sqrt[3]{x+1}

Lösung

\int_{0}^{7} 4 x 3 x + 1 d x

\frac{1209}{7}

Dezimale

172.71428 \dots

Schritte zur Lösung

\int_{0}^{7} 4 x 3 x + 1 d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{7} x 3 x + 1 d x

Wende U-Substitution an

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

Multipliziere aus

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

= 4 \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

= 4 (\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}

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

Vereinfache

4 (\frac{381}{7} - \frac{45}{4}) : \frac{1209}{7}

= \frac{1209}{7}

\int_{0}^{3} (x^{2} - 5 x + 9) d x

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