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

Lösung

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

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

Dezimale

1.14779 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{4}} sec^{3} (v) d v

Wende integrale Reduktion an

= [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} sec (v) d v

\int_{0}^{\frac{π}{4}} sec (v) d v = ln (1 + 2)

= [\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Vereinfache

[\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2) : [\frac{1}{2} sec (v) tan (v)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

= [\frac{1}{2} sec (v) tan (v)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)

Berechne die Grenzen:

\frac{1}{2}

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

Vereinfache

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

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