integral from-pi to pi of (x+1)^2sin(x)

Lösung

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

4 π

Dezimale

12.56637 \dots

Schritte zur Lösung

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

Multipliziere mit der Konjugation von

x + 1 : \frac{x ^{2} - 1}{x - 1}

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

Bestimme unbestimmte (Singularitäts-)Punkte:

x = 1

Wenn

b, a < b < c, f (b) =

unbestimmt

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

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

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

= - 2 cos (1) - π^{2} + 2 π + 4 sin (1) + 1 + π^{2} + 2 π + 2 cos (1) - 4 sin (1) - 1

Vereinfache

- 2 cos (1) - π^{2} + 2 π + 4 sin (1) + 1 + π^{2} + 2 π + 2 cos (1) - 4 sin (1) - 1 : 4 π

= 4 π

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