integral from 1 to 4 of 4xsqrt(x+1)

Lösung

\int_{1}^{4} 4 x x + 1 d x

4 (105 + \frac{- 4 2 - 50 5}{15})

Dezimale

58.11998 \dots

Schritte zur Lösung

\int_{1}^{4} 4 x x + 1 d x

Entferne die Konstante:

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

= 4 \cdot \int_{1}^{4} x x + 1 d x

Wende U-Substitution an

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

Multipliziere aus

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

= 4 \cdot \int_{2}^{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

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

\int_{2}^{5} u^{\frac{3}{2}} d u = 105 - \frac{8 2}{5}

\int_{2}^{5} u d u = \frac{10 5 - 4 2}{3}

= 4 (105 - \frac{8 2}{5} - \frac{10 5 - 4 2}{3})

Füge

105 - \frac{8 2}{5} - \frac{10 5 - 4 2}{3}

zusammen:

105 + \frac{- 4 2 - 50 5}{15}

= 4 (105 + \frac{- 4 2 - 50 5}{15})

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