integral from 0 to 1/2 of 5tcos(pit)

Lösung

\int_{0}^{\frac{1}{2}} 5 t cos (π t) d t

5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

Dezimale

0.28916 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{2}} 5 t cos (π t) d t

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{\frac{1}{2}} t cos (π t) d t

Wende die partielle Integration an

= 5 [\frac{1}{π} (t sin (π t) - π \cdot \int \frac{1}{π} sin (π t) d t)]_{0}^{\frac{1}{2}}

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

= 5 [\frac{1}{π} (t sin (π t) - π (- \frac{1}{π ^{2}} cos (π t)))]_{0}^{\frac{1}{2}}

Vereinfache

5 [\frac{1}{π} (t sin (π t) - π (- \frac{1}{π ^{2}} cos (π t)))]_{0}^{\frac{1}{2}} : 5 [\frac{1}{π} (t sin (π t) + \frac{1}{π} cos (π t))]_{0}^{\frac{1}{2}}

= 5 [\frac{1}{π} (t sin (π t) + \frac{1}{π} cos (π t))]_{0}^{\frac{1}{2}}

Berechne die Grenzen:

\frac{1}{2 π} - \frac{1}{π ^{2}}

= 5 (\frac{1}{2 π} - \frac{1}{π ^{2}})

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