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

Lösung

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

5 + \frac{1}{2} ln (2 + 5)

Dezimale

2.95788 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

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

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 5)

Berechne die Grenzen:

5

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

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