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

Lösung

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

\frac{- ln ( 2 + 5 ) + 18 5}{64}

Dezimale

0.60633 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (2)} \frac{1}{8} sec^{3} (u) tan^{2} (u) d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} sec^{3} (u) (- 1 + sec^{2} (u)) d u

Multipliziere aus

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (2)} - sec^{3} (u) + sec^{5} (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}{8} (- \int_{0}^{a r c t a n (2)} sec^{3} (u) d u + \int_{0}^{a r c t a n (2)} sec^{5} (u) d u)

\int_{0}^{a r c t a n (2)} sec^{3} (u) d u = 5 + \frac{1}{2} ln (2 + 5)

\int_{0}^{a r c t a n (2)} sec^{5} (u) d u = \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4}

= \frac{1}{8} (- (5 + \frac{1}{2} ln (2 + 5)) + \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4})

Vereinfache

\frac{1}{8} (- (5 + \frac{1}{2} ln (2 + 5)) + \frac{13 5 + \frac{3}{2} ln ( 2 + 5 )}{4}) : \frac{- ln ( 2 + 5 ) + 18 5}{64}

= \frac{- ln ( 2 + 5 ) + 18 5}{64}

\int_{0}^{2} (x^{3} - 2 x) d x

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