integral from 12 to 16 of sqrt(y^2-6)

Lösung

\int_{12}^{16} y^{2} - 6 d y

6 (- \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26)) + \frac{20 10}{3} - 138)

Dezimale

55.13000 \dots

Schritte zur Lösung

\int_{12}^{16} y^{2} - 6 d y

Trigonometrische Substitution anwenden

= \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} 6 tan^{2} (u) sec (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 6 \cdot \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 6 \cdot \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} - sec (u) + sec^{3} (u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (- \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} sec (u) d u + \int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} sec^{3} (u) d u)

\int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} sec (u) d u = ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26)

\int_{arcsec (26)}^{arcsec (\frac{8 2}{3})} sec^{3} (u) d u = \frac{20 10}{3} - 138 + \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26))

= 6 (- (ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26)) + \frac{20 10}{3} - 138 + \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26)))

Addiere gleiche Elemente:

- (ln (\frac{5 15 + 8 6}{3}) - ln (26 + 23)) + \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (26 + 23)) = - \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (26 + 23))

= 6 (- \frac{1}{2} (ln (\frac{5 15 + 8 6}{3}) - ln (23 + 26)) + \frac{20 10}{3} - 138)

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