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人気のある微分積分 >

integral from 0.25 to 0.6 of 5(1-x)^4

解

\int_{0.25}^{0.6} 5 (1 - x)^{4} d x

解

0.2270646875

解答ステップ

\int_{0.25}^{0.6} 5 (1 - x)^{4} d x

定数を除く:

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

= 5 \cdot \int_{0.25}^{0.6} (1 - x)^{4} d x

u置換積分法を適用する

= 5 \cdot \int_{0.75}^{0.4} - u^{4} d u

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

= 5 (- \int_{0.4}^{0.75} - u^{4} d u)

定数を除く:

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

= 5 (- (- \int_{0.4}^{0.75} u^{4} d u))

べき乗則を適用する

= 5 (- (- [\frac{u ^{5}}{5}]_{0.4}^{0.75}))

簡素化

= 5 [0.2 u^{5}]_{0.4}^{0.75}

限度を計算する:

0.0454129375

= 5 \cdot 0.0454129375

簡素化

= 0.2270646875

グラフ

人気の例

integral from-3 to 0 of x^3+5x^2+6x

\int_{- 3}^{0} x^{3} + 5 x^{2} + 6 x d x

integral from 0 to 0.4 of 1/(1+x^4)

\int_{0}^{0.4} \frac{1}{1 + x ^{4}} d x

integral from 0 to t of xe^{x^2}

\int_{0}^{t} x e^{x^{2}} d x

integral from 0 to 3 of te^{-st}

\int_{0}^{3} t e^{- s t} d t

integral from 0 to 9 of ((10)/((1+x)^2))

\int_{0}^{9} (\frac{10}{( 1 + x ) ^{2}}) d x