integral from 1 to 4 of 2xsqrt(x^2+1)

Lösung

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

\frac{2 ( 17 17 - 2 2 )}{3}

Dezimale

44.84291 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

= 2 \cdot \int_{1}^{4} x x^{2} + 1 d x

Wende U-Substitution an

= 2 \cdot \int_{2}^{17} u^{2} d u

Wende die Potenzregel an

= 2 [\frac{u ^{3}}{3}]_{2}^{17}

Berechne die Grenzen:

\frac{17 17 - 2 2}{3}

= 2 \cdot \frac{17 17 - 2 2}{3}

Vereinfache

= \frac{2 ( 17 17 - 2 2 )}{3}

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

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