integral from 1 to 5 of xsqrt(x-1)

Lösung

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

\frac{272}{15}

Dezimale

18.13333 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{4} (u + 1) u d u

Multipliziere aus

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

= \int_{0}^{4} 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}^{4} u^{\frac{3}{2}} d u + \int_{0}^{4} u d u

\int_{0}^{4} u^{\frac{3}{2}} d u = \frac{64}{5}

\int_{0}^{4} u d u = \frac{16}{3}

= \frac{64}{5} + \frac{16}{3}

Vereinfache

\frac{64}{5} + \frac{16}{3} : \frac{272}{15}

= \frac{272}{15}

\int_{0}^{1} \frac{( e ^{x} + 2 )}{e ^{x}} d x

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

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

\int_{0}^{1} \frac{5}{7 + 7 x ^{2}} d x

\int_{0}^{4} (1 - x) d x