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

Lösung

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

\frac{1}{2} (465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

Dezimale

15.33969 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int_{a r c t a n (2)}^{a r c t a n (8)} \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_{a r c t a n (2)}^{a r c t a n (8)} sec^{3} (u) d u

Wende integrale Reduktion an

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

\int_{a r c t a n (2)}^{a r c t a n (8)} sec (u) d u = ln (8 + 65) - ln (2 + 5)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

Vereinfache

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (2)}^{a r c t a n (8)} + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

Berechne die Grenzen:

465 - 5

= \frac{1}{2} (465 - 5 + \frac{1}{2} (ln (8 + 65) - ln (2 + 5)))

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