integral from 0 to 1/2 of-7arccos(x)

解

\int_{0}^{\frac{1}{2}} - 7 arccos (x) d x

- 7 (\frac{π}{6} - \frac{3}{2} + 1)

十進法表記

- 4.60301 \dots

解答ステップ

\int_{0}^{\frac{1}{2}} - 7 arccos (x) d x

定数を除く:

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

= - 7 \cdot \int_{0}^{\frac{1}{2}} arccos (x) d x

部分積分を適用する

= - 7 [x arccos (x) - \int - \frac{x}{1 - x ^{2}} d x]_{0}^{\frac{1}{2}}

\int - \frac{x}{1 - x ^{2}} d x = 1 - x^{2}

= - 7 [x arccos (x) - 1 - x^{2}]_{0}^{\frac{1}{2}}

限度を計算する:

\frac{π}{6} - \frac{3}{2} + 1

= - 7 (\frac{π}{6} - \frac{3}{2} + 1)