integral from 1 to 2 of x^2cos(2x)

Lösung

\int_{1}^{2} x^{2} cos (2 x) d x

\frac{7}{4} sin (4) + cos (4) - \frac{1}{4} sin (2) - \frac{1}{2} cos (2)

Dezimale

- 1.99729 \dots

Schritte zur Lösung

\int_{1}^{2} x^{2} cos (2 x) d x

Wende die partielle Integration an

= [\frac{1}{2} x^{2} sin (2 x) - \int x sin (2 x) d x]_{1}^{2}

\int x sin (2 x) d x = - \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)

= [\frac{1}{2} x^{2} sin (2 x) - (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x))]_{1}^{2}

Vereinfache

= [\frac{1}{2} x^{2} sin (2 x) + \frac{1}{2} x cos (2 x) - \frac{1}{4} sin (2 x)]_{1}^{2}

Berechne die Grenzen:

\frac{7}{4} sin (4) + cos (4) - \frac{1}{4} sin (2) - \frac{1}{2} cos (2)

= \frac{7}{4} sin (4) + cos (4) - \frac{1}{4} sin (2) - \frac{1}{2} cos (2)

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

\int_{2}^{4} arcsec (x) d x

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

\int_{0}^{\infty} e^{- 66 x} x d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} - cos (x) d x