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

Lösung

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

\frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{nπ 2} \frac{u cos ( \frac{u}{2} )}{π ^{2} n ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{nπ 2} u cos (\frac{u}{2}) d u

Wende U-Substitution an

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{0}^{πn} 4 v cos (v) d v

Entferne die Konstante:

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

= \frac{1}{π ^{2} n ^{2}} \cdot 4 \cdot \int_{0}^{πn} v cos (v) d v

Wende die partielle Integration an

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - \int sin (v) d v]_{0}^{πn}

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

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{πn}

Vereinfache

\frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{πn} : \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{0}^{πn}

= \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{0}^{πn}

Berechne die Grenzen:

(- 1)^{n} - 1

= \frac{4}{π ^{2} n ^{2}} ((- 1)^{n} - 1)

Vereinfache

= \frac{4 ( ( - 1 ) ^{n} - 1 )}{π ^{2} n ^{2}}

\int_{0}^{5} \frac{x ^{3}}{x ^{2} + 1} d x

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\int_{0}^{2 π} ∣ sin (x + 1) ∣ d x

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