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

Lösung

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

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

Dezimale

8.14577 \dots

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Wende integrale Reduktion an

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

\int_{\frac{π}{4}}^{a r c t a n (4)} sec (u) d u = ln (4 + 17) - ln (1 + 2)

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

Vereinfache

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

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

Berechne die Grenzen:

217 - \frac{1}{2}

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

Vereinfache

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

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