integral from 0 to 1 of k(1-y)y^n

Lösung

\int_{0}^{1} k \cdot (1 - y) \cdot y^{n} d y

k (\frac{1}{n + 1} - \frac{1}{n + 2})

Schritte zur Lösung

\int_{0}^{1} k (1 - y) y^{n} d y

Entferne die Konstante:

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

= k \cdot \int_{0}^{1} (1 - y) y^{n} d y

Multipliziere aus

(1 - y) y^{n} : y^{n} - y^{1 + n}

= k \cdot \int_{0}^{1} y^{n} - y^{1 + n} d y

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= k (\int_{0}^{1} y^{n} d y - \int_{0}^{1} y^{1 + n} d y)

\int_{0}^{1} y^{n} d y = \frac{1}{n + 1}

\int_{0}^{1} y^{1 + n} d y = \frac{1}{n + 2}

= k (\frac{1}{n + 1} - \frac{1}{n + 2})

\int_{- \frac{π}{2}}^{\frac{π}{2}} 7 sin (x) d x

\int_{3}^{7} (x - 3) (x + 6) d x

\int_{3}^{9} \frac{1}{9 + x ^{2}} d x

\int_{0}^{4} 9 + 2 x d x

\int_{0}^{1} sin^{2} (kπ x) d x