integral from 0 to pi of sin(x)-x^2+pix

Lösung

\int_{0}^{π} sin (x) - x^{2} + π x d x

\frac{π ^{3}}{6} + 2

Dezimale

7.16771 \dots

Schritte zur Lösung

\int_{0}^{π} sin (x) - x^{2} + π x d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} sin (x) d x - \int_{0}^{π} x^{2} d x + \int_{0}^{π} π x d x

\int_{0}^{π} sin (x) d x = 2

\int_{0}^{π} x^{2} d x = \frac{π ^{3}}{3}

\int_{0}^{π} π x d x = \frac{π ^{3}}{2}

= 2 - \frac{π ^{3}}{3} + \frac{π ^{3}}{2}

Ziehe Brüche zusammen

- \frac{π ^{3}}{3} + \frac{π ^{3}}{2} : \frac{π ^{3}}{6}

= \frac{π ^{3}}{6} + 2

\int_{- 2}^{2} x + 5 d x

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

\int_{0}^{L} sin^{2} (x) d x

\int_{0}^{\frac{π}{2}} (2 - sin (x))^{2} d x