integral from 0 to 7 of sqrt(1+x^2)

Lösung

\int_{0}^{7} 1 + x^{2} d x

\frac{35}{2} + \frac{1}{2} ln (7 + 52)

Dezimale

26.07079 \dots

Schritte zur Lösung

\int_{0}^{7} 1 + x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (7)} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (7)} sec (u) d u

\int_{0}^{a r c t a n (7)} sec (u) d u = ln (7 + 52)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (7)} + \frac{1}{2} ln (7 + 52)

Berechne die Grenzen:

\frac{35}{2}

= \frac{35}{2} + \frac{1}{2} ln (7 + 52)

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