integral from 0 to 7 of 8x*\sqrt[3]{x+1}

Lösung

\int_{0}^{7} 8 x \cdot 3 x + 1 d x

\frac{2418}{7}

Dezimale

345.42857 \dots

Schritte zur Lösung

\int_{0}^{7} 8 x 3 x + 1 d x

Entferne die Konstante:

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

= 8 \cdot \int_{0}^{7} x 3 x + 1 d x

Wende U-Substitution an

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

Multipliziere aus

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

= 8 \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

= 8 (\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}

= 8 (\frac{381}{7} - \frac{45}{4})

Vereinfache

8 (\frac{381}{7} - \frac{45}{4}) : \frac{2418}{7}

= \frac{2418}{7}

d er i v a t i v e f (x) = ln (x + 6)

\int_{0}^{15} 1000 e^{- 0.05 t} d t

x \to 1 lim (3 - 3 x)

y^{^{'}} + \frac{1}{6} y = 40

\frac{\partial}{\partial y} ((x + y + z)^{2})