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

Lösung

\int_{0}^{2 π} 1 + x^{2} d x

π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Dezimale

21.25629 \dots

Schritte zur Lösung

\int_{0}^{2 π} 1 + x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (2 π)} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2 π)} sec (u) d u

\int_{0}^{a r c t a n (2 π)} sec (u) d u = ln (2 π + 4 π^{2} + 1)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2 π)} + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

Berechne die Grenzen:

π 4 π^{2} + 1

= π 4 π^{2} + 1 + \frac{1}{2} ln (2 π + 4 π^{2} + 1)

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