integral from 0 to 3 of sqrt(1+4x^2)

Lösung

\int_{0}^{3} 1 + 4 x^{2} d x

\frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

Dezimale

9.74708 \dots

Schritte zur Lösung

\int_{0}^{3} 1 + 4 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (6)} \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 (6)} 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 (6)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (6)} sec (u) d u)

\int_{0}^{a r c t a n (6)} sec (u) d u = ln (6 + 37)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (6)} + \frac{1}{2} ln (6 + 37))

Berechne die Grenzen:

337

= \frac{1}{2} (337 + \frac{1}{2} ln (6 + 37))

\frac{d}{d t} (a t^{2} + b t + c)

d er i v a t i v e 2

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