integral from 0 to 3 of sqrt(1+(8x)^2)

Lösung

\int_{0}^{3} 1 + (8 x)^{2} d x

\frac{1}{8} (12577 + \frac{1}{2} ln (24 + 577))

Dezimale

36.27321 \dots

Schritte zur Lösung

\int_{0}^{3} 1 + (8 x)^{2} d x

Vereinfache

1 + (8 x)^{2} : 1 + 64 x^{2}

= \int_{0}^{3} 1 + 64 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (24)} \frac{1}{8} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{0}^{a r c t a n (24)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (24)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (24)} sec (u) d u)

\int_{0}^{a r c t a n (24)} sec (u) d u = ln (24 + 577)

= \frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (24)} + \frac{1}{2} ln (24 + 577))

Vereinfache

\frac{1}{8} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (24)} + \frac{1}{2} ln (24 + 577)) : \frac{1}{8} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (24)} + \frac{1}{2} ln (24 + 577))

= \frac{1}{8} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (24)} + \frac{1}{2} ln (24 + 577))

Berechne die Grenzen:

12577

= \frac{1}{8} (12577 + \frac{1}{2} ln (24 + 577))