integral from 0 to 1 of x(1-x)^{193}

Lösung

\int_{0}^{1} x (1 - x)^{193} d x

\frac{1}{37830}

Dezimale

0.00002 \dots

Schritte zur Lösung

\int_{0}^{1} x (1 - x)^{193} d x

Wende U-Substitution an

= \int_{1}^{0} - (- u + 1) u^{193} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - (- u + 1) u^{193} d u

Entferne die Konstante:

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

= - (- \int_{0}^{1} (- u + 1) u^{193} d u)

Multipliziere aus

(- u + 1) u^{193} : - u^{194} + u^{193}

= - (- \int_{0}^{1} - u^{194} + u^{193} 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_{0}^{1} u^{194} d u + \int_{0}^{1} u^{193} d u))

\int_{0}^{1} u^{194} d u = \frac{1}{195}

\int_{0}^{1} u^{193} d u = \frac{1}{194}

= - (- (- \frac{1}{195} + \frac{1}{194}))

Vereinfache

- (- (- \frac{1}{195} + \frac{1}{194})) : \frac{1}{37830}

= \frac{1}{37830}

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