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

Lösung

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

17 + \frac{1}{4} ln (4 + 17)

Dezimale

4.64678 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (4)} \frac{1}{2} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (4)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (4)} sec (u) d u)

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Berechne die Grenzen:

217

= \frac{1}{2} (217 + \frac{1}{2} ln (4 + 17))

Vereinfache

= 17 + \frac{1}{4} ln (4 + 17)

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