integral from 0 to 1 of 2xsqrt(x+1)

Lösung

\int_{0}^{1} 2 x x + 1 d x

2 \cdot \frac{4 2 + 4}{15}

Dezimale

1.28758 \dots

Schritte zur Lösung

\int_{0}^{1} 2 x x + 1 d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} x x + 1 d x

Wende U-Substitution an

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

Multipliziere aus

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

= 2 \cdot \int_{1}^{2} 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

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

\int_{1}^{2} u^{\frac{3}{2}} d u = \frac{8 2 - 2}{5}

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

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

Füge

\frac{8 2 - 2}{5} - \frac{4 2 - 2}{3}

zusammen:

\frac{4 2 + 4}{15}

= 2 \cdot \frac{4 2 + 4}{15}

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