integral from 0 to 80 of (1-t/(80))^a

解

\int_{0}^{80} (1 - \frac{t}{80})^{a} d t

\frac{80}{a + 1}

解答ステップ

\int_{0}^{80} (1 - \frac{t}{80})^{a} d t

u置換積分法を適用する

= \int_{0}^{1} (1 - u)^{a} \cdot 80 d u

定数を除く:

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

= 80 \cdot \int_{0}^{1} (1 - u)^{a} d u

u置換積分法を適用する

= 80 \cdot \int_{1}^{0} - v^{a} d v

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

= 80 (- \int_{0}^{1} - v^{a} d v)

定数を除く:

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

= 80 (- (- \int_{0}^{1} v^{a} d v))

べき乗則を適用する

= 80 (- (- [\frac{v ^{a + 1}}{a + 1}]_{0}^{1}))

簡素化

= 80 [\frac{v ^{a + 1}}{a + 1}]_{0}^{1}

限度を計算する:

\frac{1}{a + 1}

= 80 \cdot \frac{1}{a + 1}

簡素化

= \frac{80}{a + 1}