integral from 0 to 2 of 2tsin(2pit)

Lösung

\int_{0}^{2} 2 t sin (2 π t) d t

- \frac{2}{π}

Dezimale

- 0.63661 \dots

Schritte zur Lösung

\int_{0}^{2} 2 t sin (2 π t) d t

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{2} t sin (2 π t) d t

Wende die partielle Integration an

= 2 [- \frac{1}{2 π} t cos (2 π t) - \int - \frac{1}{2 π} cos (2 π t) d t]_{0}^{2}

\int - \frac{1}{2 π} cos (2 π t) d t = - \frac{1}{4 π ^{2}} sin (2 π t)

= 2 [- \frac{1}{2 π} t cos (2 π t) - (- \frac{1}{4 π ^{2}} sin (2 π t))]_{0}^{2}

Vereinfache

= 2 [- \frac{1}{2 π} t cos (2 π t) + \frac{1}{4 π ^{2}} sin (2 π t)]_{0}^{2}

Berechne die Grenzen:

- \frac{1}{π}

= 2 (- \frac{1}{π})

Vereinfache

= - \frac{2}{π}

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