integral from 0 to pi of sqrt(sin(e))

解

\int_{0}^{π} sin (e) d e

- 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - π) ∣) + 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - 0) ∣)

解答ステップ

\int_{0}^{π} sin (e) d e

非初等積分を使用する:

\int sin (x) d x = - 2 E (\frac{1}{2} (\frac{π}{2} - x) ∣ 2)

= [- 2 E (\frac{1}{2} (\frac{π}{2} - e) ∣ 2)]_{0}^{π}

限度を計算する:

- 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - π) ∣) + 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - 0) ∣)

= - 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - π) ∣) + 2 E (2 \cdot \frac{1}{2} (\frac{π}{2} - 0) ∣)