integral from 1 to 2 of 15sqrt(4x^2-3)

Lösung

\int_{1}^{2} 15 4 x^{2} - 3 d x

\frac{15 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

Dezimale

36.11777 \dots

Schritte zur Lösung

\int_{1}^{2} 15 4 x^{2} - 3 d x

Entferne die Konstante:

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

= 15 \cdot \int_{1}^{2} 4 x^{2} - 3 d x

Trigonometrische Substitution anwenden

= 15 \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} \frac{3}{2} tan^{2} (u) sec (u) d u

Entferne die Konstante:

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

= 15 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} tan^{2} (u) sec (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 15 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} (- 1 + sec^{2} (u)) sec (u) d u

Multipliziere aus

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

= 15 \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{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

= 15 \cdot \frac{3}{2} (- \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (u) d u + \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (u) d u)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (u) d u = ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (u) d u = \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))

= 15 \cdot \frac{3}{2} (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)))

Vereinfache

15 \cdot \frac{3}{2} (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))) : \frac{15 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

= \frac{15 ( 3 ln ( 3 ) + 4 13 - 3 ln ( 4 + 13 ) - 2 )}{4}

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\frac{\partial}{\partial y} (e^{- y} cos (π x))