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

Lösung

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

\frac{1}{8} (4 (221 + 3) + 10 ln (\frac{4}{5} + \frac{21}{5}) + 5 ln (5))

Dezimale

8.76974 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{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_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} \cdot \int_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} sec (u) d u

\int_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} sec (u) d u = ln (\frac{4}{5} + \frac{21}{5}) + \frac{1}{2} ln (5)

= \frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} (ln (\frac{4}{5} + \frac{21}{5}) + \frac{1}{2} ln (5))

Vereinfache

\frac{5}{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (\frac{2}{5})}^{a r c t a n (\frac{4}{5})} + \frac{1}{2} (ln (\frac{4}{5} + \frac{21}{5}) + \frac{1}{2} ln (5)) : \frac{1}{8} 20 [\frac{sec ( u ) tan ( u )}{2}]_{- a r c t a n (\frac{2 5}{5})}^{a r c t a n (\frac{4 5}{5})} + 10 ln (\frac{4}{5} + \frac{21}{5}) + 5 ln (5)

= \frac{1}{8} 20 [\frac{sec ( u ) tan ( u )}{2}]_{- a r c t a n (\frac{2 5}{5})}^{a r c t a n (\frac{4 5}{5})} + 10 ln (\frac{4}{5} + \frac{21}{5}) + 5 ln (5)

Berechne die Grenzen:

\frac{2 21}{5} + \frac{3}{5}

= \frac{1}{8} (20 (\frac{2 21}{5} + \frac{3}{5}) + 10 ln (\frac{4}{5} + \frac{21}{5}) + 5 ln (5))

Vereinfache

= \frac{1}{8} (4 (221 + 3) + 10 ln (\frac{4}{5} + \frac{21}{5}) + 5 ln (5))

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