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

Lösung

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

\frac{1}{4} (217 + \frac{1}{2} ln (4 + 17))

Dezimale

2.32339 \dots

Schritte zur Lösung

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

Vereinfache

1 + (4 x)^{2} : 1 + 16 x^{2}

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

Trigonometrische Substitution anwenden

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

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= \frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Vereinfache

\frac{1}{4} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

= \frac{1}{4} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

Berechne die Grenzen:

217

= \frac{1}{4} (217 + \frac{1}{2} ln (4 + 17))

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