integral from 0 to 1 of x(1-x)^2

Lösung

\int_{0}^{1} x (1 - x)^{2} d x

\frac{1}{12}

Dezimale

0.08333 \dots

Schritte zur Lösung

\int_{0}^{1} x (1 - x)^{2} d x

Multipliziere aus

x (1 - x)^{2} : x - 2 x^{2} + x^{3}

= \int_{0}^{1} x - 2 x^{2} + x^{3} 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_{0}^{1} x d x - \int_{0}^{1} 2 x^{2} d x + \int_{0}^{1} x^{3} d x

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 2 x^{2} d x = \frac{2}{3}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= \frac{1}{2} - \frac{2}{3} + \frac{1}{4}

Vereinfache

\frac{1}{2} - \frac{2}{3} + \frac{1}{4} : \frac{1}{12}

= \frac{1}{12}

x \to + (- 2) lim (x^{2} - x + 4)

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 t e^{3 t}

d er i v a t i v e f (x) = ln (9 sin^{2} (x))

\int \frac{3 x - 1}{x ^{2} - 4} d x

l a pl a ce t r an s f or m - 3 (sin (4 (t - \frac{5 π}{4})))