integral from 0 to pi/2 of x^2cos(4x)

Lösung

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

\frac{π}{16}

Dezimale

0.19634 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

Wende U-Substitution an

= \int_{0}^{2 π} \frac{u ^{2} cos ( u )}{64} d u

Entferne die Konstante:

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

= \frac{1}{64} \cdot \int_{0}^{2 π} u^{2} cos (u) d u

Wende die partielle Integration an

= \frac{1}{64} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{2 π}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{64} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{2 π}

Berechne die Grenzen:

4 π

= \frac{1}{64} \cdot 4 π

Vereinfache

= \frac{π}{16}

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