integral from 0 to 1 of sqrt(7+4t^2)

Lösung

\int_{0}^{1} 7 + 4 t^{2} d t

\frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

Dezimale

2.87960 \dots

Schritte zur Lösung

\int_{0}^{1} 7 + 4 t^{2} d t

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (\frac{2}{7})} \frac{7}{2} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{7}{2} \cdot \int_{0}^{a r c t a n (\frac{2}{7})} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{0}^{a r c t a n (\frac{2}{7})} sec (u) d u = ln (\frac{2}{7} + \frac{11}{7})

= \frac{7}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7})

Vereinfache

\frac{7}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + \frac{1}{2} ln (\frac{2}{7} + \frac{11}{7}) : \frac{1}{4} 14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7})

= \frac{1}{4} 14 [\frac{sec ( u ) tan ( u )}{2}]_{0}^{a r c t a n (\frac{2}{7})} + 7 ln (\frac{2}{7} + \frac{11}{7})

Berechne die Grenzen:

\frac{11}{7}

= \frac{1}{4} (14 \cdot \frac{11}{7} + 7 ln (\frac{2}{7} + \frac{11}{7}))

Vereinfache

= \frac{1}{4} (211 + 7 ln (\frac{2}{7} + \frac{11}{7}))

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