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

Lösung

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

\frac{25 5 + 1}{15}

Dezimale

3.79344 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{5} \frac{u ( u - 1 )}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{1}^{5} u (u - 1) d u

Multipliziere aus

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

= \frac{1}{4} \cdot \int_{1}^{5} 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

= \frac{1}{4} (\int_{1}^{5} u^{\frac{3}{2}} d u - \int_{1}^{5} u d u)

\int_{1}^{5} u^{\frac{3}{2}} d u = 105 - \frac{2}{5}

\int_{1}^{5} u d u = \frac{10 5 - 2}{3}

= \frac{1}{4} (105 - \frac{2}{5} - \frac{10 5 - 2}{3})

Vereinfache

\frac{1}{4} (105 - \frac{2}{5} - \frac{10 5 - 2}{3}) : \frac{25 5 + 1}{15}

= \frac{25 5 + 1}{15}

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