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

Lösung

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

\frac{1209}{28}

Dezimale

43.17857 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

= \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

= \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}

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

Vereinfache

\frac{381}{7} - \frac{45}{4} : \frac{1209}{28}

= \frac{1209}{28}

f (x) = x^{3 x}

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (7) = 7

\int x^{5} sin (x^{6} + 2) d x

x \to π lim (cot (x))

\int \frac{2 x ^{2} + 1}{( x - 3 ) ^{3}} d x