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

解

\int_{π}^{0} (sin (x) - 2)^{2} d x

- \frac{π}{2} + 8 - 4 π

十進法表記

- 6.13716 \dots

解答ステップ

\int_{π}^{0} (sin (x) - 2)^{2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

拡張

(sin (x) - 2)^{2} : sin^{2} (x) - 4 sin (x) + 4

= - \int_{0}^{π} sin^{2} (x) - 4 sin (x) + 4 d x

総和規則を適用する:

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

= - (\int_{0}^{π} sin^{2} (x) d x - \int_{0}^{π} 4 sin (x) d x + \int_{0}^{π} 4 d x)

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

\int_{0}^{π} 4 sin (x) d x = 8

\int_{0}^{π} 4 d x = 4 π

= - (\frac{π}{2} - 8 + 4 π)

簡素化

= - \frac{π}{2} + 8 - 4 π