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

Lösung

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

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

Dezimale

1.42925 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Trigonometrische Substitution anwenden

= 2 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} 16 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

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) tan^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{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)

= 2 \cdot 16 \cdot \int_{0}^{a r c t a n (\frac{1}{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

= 2 \cdot 16 (- \int_{0}^{a r c t a n (\frac{1}{2})} sec^{3} (u) d u + \int_{0}^{a r c t a n (\frac{1}{2})} sec^{5} (u) d u)

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

\int_{0}^{a r c t a n (\frac{1}{2})} sec^{5} (u) d u = \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64}

= 2 \cdot 16 - (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2})) + \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64}

Vereinfache

2 \cdot 16 - (\frac{5}{8} + \frac{1}{2} ln (\frac{1 + 5}{2})) + \frac{11 5 + 24 ln ( \frac{1 + 5}{2} )}{64} : \frac{- 8 ln ( \frac{1 + 5}{2} ) + 3 5}{2}

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

\int e^{- 0.02 x} d x

\int \frac{1}{( 2 x - 1 )} d x

t^{2} \frac{d y}{d t} - t = 1 + y + y t

y^{^{''}} = ey

\frac{d}{d x} (e^{x + 5})