integral from 1 to 2 of xsqrt(x+1)

Lösung

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

- 23 + \frac{- 4 2 + 54 3}{15}

Dezimale

2.39415 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{3} (u - 1) u d u

Multipliziere aus

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

= \int_{2}^{3} 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_{2}^{3} u^{\frac{3}{2}} d u - \int_{2}^{3} u d u

\int_{2}^{3} u^{\frac{3}{2}} d u = \frac{18 3 - 8 2}{5}

\int_{2}^{3} u d u = 23 - \frac{4 2}{3}

= \frac{18 3 - 8 2}{5} - (23 - \frac{4 2}{3})

Vereinfache

\frac{18 3 - 8 2}{5} - (23 - \frac{4 2}{3}) : - 23 + \frac{- 4 2 + 54 3}{15}

= - 23 + \frac{- 4 2 + 54 3}{15}

\int_{1}^{2} x^{6} d x

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

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

\int_{10}^{y} \frac{10}{x ^{2}} d x

\int_{0}^{π} 3 x sin (2 x) d x