integral from-1 to 1 of t*sin(npit)

Lösung

\int_{- 1}^{1} t \cdot sin (nπ t) d t

- \frac{2 ( - 1 ) ^{n}}{πn}

Schritte zur Lösung

\int_{- 1}^{1} t sin (nπ t) d t

Wende die partielle Integration an

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

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

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

Vereinfache

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

Berechne die Grenzen:

- (- 1)^{n} \frac{2}{πn}

= - (- 1)^{n} \frac{2}{πn}

Vereinfache

= - \frac{2 ( - 1 ) ^{n}}{πn}

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