integral from 0 to 20 of sqrt(1+4x^2)

Lösung

\int_{0}^{20} 1 + 4 x^{2} d x

\frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

Dezimale

401.22052 \dots

Schritte zur Lösung

\int_{0}^{20} 1 + 4 x^{2} d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (40)} \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 (40)} 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 (40)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (40)} sec (u) d u)

\int_{0}^{a r c t a n (40)} sec (u) d u = ln (40 + 1601)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (40)} + \frac{1}{2} ln (40 + 1601))

Berechne die Grenzen:

201601

= \frac{1}{2} (201601 + \frac{1}{2} ln (40 + 1601))

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