integral from 0 to 1 of sqrt(5+4t^2)

Lösung

\int_{0}^{1} 5 + 4 t^{2} d t

\frac{1}{8} (12 + 5 ln (5))

Dezimale

2.50589 \dots

Schritte zur Lösung

\int_{0}^{1} 5 + 4 t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (\frac{2}{5})} \frac{5}{2} sec^{3} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{5}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{5})} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{0}^{a r c t a n (\frac{2}{5})} sec (u) d u = \frac{1}{2} ln (5)

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

Vereinfache

\frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + \frac{1}{2} \cdot \frac{1}{2} ln (5) : \frac{1}{8} 20 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + 5 ln (5)

= \frac{1}{8} 20 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{5})} + 5 ln (5)

Berechne die Grenzen:

\frac{3}{5}

= \frac{1}{8} (20 \cdot \frac{3}{5} + 5 ln (5))

Vereinfache

= \frac{1}{8} (12 + 5 ln (5))

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