integral from 0 to 2 of x^3sqrt(4x^2+1)

Lösung

\int_{0}^{2} x^{3} 4 x^{2} + 1 d x

\frac{391 17 + 1}{120}

Dezimale

13.44278 \dots

Schritte zur Lösung

\int_{0}^{2} x^{3} 4 x^{2} + 1 d x

Wende U-Substitution an

= \int_{1}^{17} \frac{u ( u - 1 )}{32} d u

Entferne die Konstante:

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

= \frac{1}{32} \cdot \int_{1}^{17} u (u - 1) d u

Multipliziere aus

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

= \frac{1}{32} \cdot \int_{1}^{17} 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}{32} (\int_{1}^{17} u^{\frac{3}{2}} d u - \int_{1}^{17} u d u)

\int_{1}^{17} u^{\frac{3}{2}} d u = \frac{578 17 - 2}{5}

\int_{1}^{17} u d u = \frac{34 17 - 2}{3}

= \frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3})

Vereinfache

\frac{1}{32} (\frac{578 17 - 2}{5} - \frac{34 17 - 2}{3}) : \frac{391 17 + 1}{120}

= \frac{391 17 + 1}{120}

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