integral from 0 to 1 of sin(npix)

Lösung

\int_{0}^{1} sin (n \cdot π \cdot x) d x

\frac{- ( - 1 ) ^{n} + 1}{πn}

Schritte zur Lösung

\int_{0}^{1} sin (nπ x) d x

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{πn} \cdot \int_{0}^{πn} sin (u) d u

Nutze das gemeinsame Integral :

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

= \frac{1}{πn} [- cos (u)]_{0}^{πn}

Berechne die Grenzen:

- (- 1)^{n} + 1

= \frac{1}{πn} (- (- 1)^{n} + 1)

Vereinfache

= \frac{- ( - 1 ) ^{n} + 1}{πn}

\int_{1}^{2} 20 x^{3} + \frac{6}{x ^{2}} d x

\int_{3 x}^{2} \frac{t ^{2} - 1}{t ^{2} + 1} d t

\int_{1}^{3} 3 x^{2} - 2 d x

\int_{0}^{1} 2 t e^{- t^{2}} d t

\int_{1}^{4} (x - x + 2) d x