integral from (-1)/8 to 1/8 of sin(2pix)

Lösung

\int_{\frac{- 1}{8}}^{\frac{1}{8}} sin (2 π x) d x

0

Schritte zur Lösung

\int_{\frac{- 1}{8}}^{\frac{1}{8}} sin (2 π x) d x

Wende U-Substitution an

= \int_{- \frac{π}{4}}^{\frac{π}{4}} sin (u) \frac{1}{2 π} d u

Entferne die Konstante:

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

= \frac{1}{2 π} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} sin (u) d u

Nutze das gemeinsame Integral :

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

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

Berechne die Grenzen:

0

= \frac{1}{2 π} \cdot 0

Vereinfache

= 0

\int_{0}^{π} x \cdot cos (n x) d x

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\int_{2}^{3} 4 x e^{x^{2} - 5} d x

\int_{1}^{5} x 5 x^{2} - 4 d x