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

Lösung

\int_{0}^{4} x (x + 2) d x

- \frac{1}{2} ln (26 + 5) + 56

Dezimale

11.10123 \dots

Schritte zur Lösung

\int_{0}^{4} x (x + 2) d x

Vervollständige das Quadrat

x (x + 2) : (x + 1)^{2} - 1

= \int_{0}^{4} (x + 1)^{2} - 1 d x

Wende U-Substitution an

= \int_{1}^{5} u^{2} - 1 d u

Trigonometrische Substitution anwenden

= \int_{0}^{arcsec (5)} tan^{2} (v) sec (v) d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{arcsec (5)} (- 1 + sec^{2} (v)) sec (v) d v

Multipliziere aus

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

= \int_{0}^{arcsec (5)} - sec (v) + sec^{3} (v) d v

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}^{arcsec (5)} sec (v) d v + \int_{0}^{arcsec (5)} sec^{3} (v) d v

\int_{0}^{arcsec (5)} sec (v) d v = ln (26 + 5)

\int_{0}^{arcsec (5)} sec^{3} (v) d v = 56 + \frac{1}{2} ln (26 + 5)

= - ln (26 + 5) + 56 + \frac{1}{2} ln (26 + 5)

Vereinfache

- ln (26 + 5) + 56 + \frac{1}{2} ln (26 + 5) : - \frac{1}{2} ln (26 + 5) + 56

= - \frac{1}{2} ln (26 + 5) + 56

\int_{0}^{1} 6 cos (4 x) d x

\frac{d}{d x} (arcsin (e^{x}))

\int \frac{x ^{2}}{5 - x} d x

d er i v a t i v e (2 z + e^{- z^{2}})^{6}

f (x) = 2 - x