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

Lösung

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

65 - \frac{17}{2} + \frac{1}{8} ln (8 + 65) - \frac{1}{8} ln (4 + 17)

Dezimale

6.08592 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{a r c t a n (4)}^{a r c t a n (8)} \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_{a r c t a n (4)}^{a r c t a n (8)} sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{a r c t a n (4)}^{a r c t a n (8)} sec (u) d u)

\int_{a r c t a n (4)}^{a r c t a n (8)} sec (u) d u = ln (8 + 65) - ln (4 + 17)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (4 + 17)))

Vereinfache

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (4)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (4 + 17))) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (4)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (4 + 17)))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (4)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (4 + 17)))

Berechne die Grenzen:

465 - 217

= \frac{1}{4} (465 - 217 + \frac{1}{2} (ln (8 + 65) - ln (4 + 17)))

Vereinfache

= 65 - \frac{17}{2} + \frac{1}{8} ln (8 + 65) - \frac{1}{8} ln (4 + 17)

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