integral from 0 to 3 of sqrt(1+16x^2)

Lösung

\int_{0}^{3} 1 + 16 x^{2} d x

\frac{1}{4} (6145 + \frac{1}{2} ln (12 + 145))

Dezimale

18.45986 \dots

Schritte zur Lösung

\int_{0}^{3} 1 + 16 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (12)} \frac{1}{4} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{a r c t a n (12)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (12)} sec (u) d u)

\int_{0}^{a r c t a n (12)} sec (u) d u = ln (12 + 145)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

Vereinfache

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (12)} + \frac{1}{2} ln (12 + 145))

Berechne die Grenzen:

6145

= \frac{1}{4} (6145 + \frac{1}{2} ln (12 + 145))

in v erse l a pl a ce x^{2}

(4 x y^{2} + 3 y) d x + (4 x^{2} y + 3 x) d y = 0

in v erse l a pl a ce \frac{2250}{5 x ^{2} + x}

s l o p e (- 12) (- 1 - 3)

d er i v a t i v e 4 e^{x} cos (x)