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

Lösung

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

\frac{2 2 - 1}{3}

Dezimale

0.60947 \dots

Schritte zur Lösung

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

Vereinfache

x^{2} + x^{4} : x^{2} 1 + x^{2}

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

x^{2} = (x),

angenommen

x \geq 0

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

Wende U-Substitution an

= \int_{1}^{2} u^{2} d u

Wende die Potenzregel an

= [\frac{u ^{3}}{3}]_{1}^{2}

Berechne die Grenzen:

\frac{2 2}{3} - \frac{1}{3}

= \frac{2 2}{3} - \frac{1}{3}

Vereinfache

= \frac{2 2 - 1}{3}

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