integral from 0 to 2 of sin((npix)/2)

Lösung

\int_{0}^{2} sin (\frac{nπ x}{2}) d x

\frac{2 ( - ( - 1 ) ^{n} + 1 )}{πn}

Schritte zur Lösung

\int_{0}^{2} sin (\frac{nπ x}{2}) d x

\frac{nπ x}{2} = \frac{nπ}{2} x

= \int_{0}^{2} sin (\frac{nπ}{2} x) d x

Wende U-Substitution an

= \int_{0}^{πn} \frac{2 sin ( u )}{πn} d u

Entferne die Konstante:

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

= \frac{2}{πn} \cdot \int_{0}^{πn} sin (u) d u

Nutze das gemeinsame Integral :

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

= \frac{2}{πn} [- cos (u)]_{0}^{πn}

Berechne die Grenzen:

- (- 1)^{n} + 1

= \frac{2}{πn} (- (- 1)^{n} + 1)

Vereinfache

= \frac{2 ( - ( - 1 ) ^{n} + 1 )}{πn}

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

\int_{2}^{10} \frac{3}{5 x - 1} d x

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

\int_{1}^{3} 4 x d x

\int_{- 6}^{4} \frac{3}{x ^{4}} d x