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

Lösung

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

25 + ln (2 + 5)

Dezimale

5.91577 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Wende integrale Reduktion an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

5

= 25 + ln (2 + 5)