integral from 0 to 3 of sqrt(1+1296x^2)

Lösung

\int_{0}^{3} 1 + 1296 x^{2} d x

\frac{1}{36} (5411665 + \frac{1}{2} ln (108 + 11665))

Dezimale

162.08160 \dots

Schritte zur Lösung

\int_{0}^{3} 1 + 1296 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (108)} \frac{1}{36} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{36} \cdot \int_{0}^{a r c t a n (108)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (108)} sec (u) d u)

\int_{0}^{a r c t a n (108)} sec (u) d u = ln (108 + 11665)

= \frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

Vereinfache

\frac{1}{36} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665)) : \frac{1}{36} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

= \frac{1}{36} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (108)} + \frac{1}{2} ln (108 + 11665))

Berechne die Grenzen:

5411665

= \frac{1}{36} (5411665 + \frac{1}{2} ln (108 + 11665))

\int_{2}^{3} 3 sin (\frac{π}{3} x) d x

\int_{0}^{π} 3 x d x

\int_{- 2}^{- 1} \frac{1}{( x + 1 ) ^{\frac{4}{3}}} d x

\int_{0}^{\frac{π}{2}} 6 sin (t) d t

\int_{0.5}^{2} 0.09375 (4 - x^{2}) d x