integral from 1 to 2 of x(sqrt(x-1))

Lösung

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

\frac{16}{15}

Dezimale

1.06666 \dots

Schritte zur Lösung

\int_{1}^{2} x x - 1 d x

Wende U-Substitution an

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

Multipliziere aus

(u + 1) u : u^{\frac{3}{2}} + u

= \int_{0}^{1} u^{\frac{3}{2}} + 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_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

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

= \frac{2}{5} + \frac{2}{3}

Vereinfache

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

\int_{3}^{4} - x ln (x - 3) d x

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

\int_{0}^{2} 3 x d x

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

\int_{0}^{1} x^{2} - x^{5} d x