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

Lösung

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

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

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

Faktorisiere

1 + 2 t^{2} + t^{4} : (t^{2} + 1)^{2}

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Wende Exponentenregel an:

a^{2} = ∣ a ∣

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Entferne die Extremwerte

= \int_{0}^{1} (t^{2} + 1)^{2} d t

Wende Exponentenregel an:

a^{2} = ∣ a ∣

= \int_{0}^{1} (t^{2} + 1) d t

Entferne die Extremwerte

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{1} t^{2} d t = \frac{1}{3}

\int_{0}^{1} 1 d t = 1

= \frac{1}{3} + 1

Vereinfache

\frac{1}{3} + 1 : \frac{4}{3}

= \frac{4}{3}

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