integral from-1 to 1 of x(1+x)^4

Lösung

\int_{- 1}^{1} x (1 + x)^{4} d x

\frac{64}{15}

Dezimale

4.26666 \dots

Schritte zur Lösung

\int_{- 1}^{1} x (1 + x)^{4} d x

Multipliziere aus

x (1 + x)^{4} : x + 4 x^{2} + 6 x^{3} + 4 x^{4} + x^{5}

= \int_{- 1}^{1} x + 4 x^{2} + 6 x^{3} + 4 x^{4} + x^{5} d x

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}^{1} x d x + \int_{- 1}^{1} 4 x^{2} d x + \int_{- 1}^{1} 6 x^{3} d x + \int_{- 1}^{1} 4 x^{4} d x + \int_{- 1}^{1} x^{5} d x

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 4 x^{2} d x = \frac{8}{3}

\int_{- 1}^{1} 6 x^{3} d x = 0

\int_{- 1}^{1} 4 x^{4} d x = \frac{8}{5}

\int_{- 1}^{1} x^{5} d x = 0

= 0 + \frac{8}{3} + 0 + \frac{8}{5} + 0

Vereinfache

0 + \frac{8}{3} + 0 + \frac{8}{5} + 0 : \frac{64}{15}

= \frac{64}{15}

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

\int_{\frac{π}{2}}^{x} sin (13 t) d t

\int_{- 2}^{3} (4 - ∣ x ∣) d x

\int_{1}^{1} tan (x) d x

\int_{2}^{3} (e^{2 x} + e^{x}) d x