integral from 1 to 2 of pi(x^{-1251})^2

解

\int_{1}^{2} π (x^{- 1251})^{2} d x

π (- \frac{1}{2 ^{2501} \cdot 2501} + \frac{1}{2501})

解答ステップ

\int_{1}^{2} π (x^{- 1251})^{2} d x

定数を除く:

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

= π \cdot \int_{1}^{2} (x^{- 1251})^{2} d x

簡素化

(x^{- 1251})^{2} : \frac{1}{x ^{2502}}

= π \cdot \int_{1}^{2} \frac{1}{x ^{2502}} d x

べき乗則を適用する

= π [- \frac{1}{2501 x ^{2501}}]_{1}^{2}

限度を計算する:

- \frac{1}{2 ^{2501} \cdot 2501} + \frac{1}{2501}

= π (- \frac{1}{2 ^{2501} \cdot 2501} + \frac{1}{2501})