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人気のある微分積分 >

integral from 0 to 1 of nx(1-x^2)^n

解

\int_{0}^{1} n x (1 - x^{2})^{n} d x

解

\frac{n}{2 ( n + 1 )}

解答ステップ

\int_{0}^{1} n x (1 - x^{2})^{n} d x

定数を除く:

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

= n \cdot \int_{0}^{1} x (1 - x^{2})^{n} d x

u置換積分法を適用する

= n \cdot \int_{0}^{1} \frac{( 1 - u ) ^{n}}{2} d u

定数を除く:

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

= n \frac{1}{2} \cdot \int_{0}^{1} (1 - u)^{n} d u

u置換積分法を適用する

= n \frac{1}{2} \cdot \int_{1}^{0} - v^{n} d v

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

= n \frac{1}{2} (- \int_{0}^{1} - v^{n} d v)

定数を除く:

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

= n \frac{1}{2} (- (- \int_{0}^{1} v^{n} d v))

べき乗則を適用する

= n \frac{1}{2} (- (- [\frac{v ^{n + 1}}{n + 1}]_{0}^{1}))

簡素化

= \frac{1}{2} n [\frac{v ^{n + 1}}{n + 1}]_{0}^{1}

限度を計算する:

\frac{1}{n + 1}

= \frac{1}{2} n \frac{1}{n + 1}

簡素化

= \frac{n}{2 ( n + 1 )}

人気の例

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\int_{\frac{π}{4}}^{\frac{π}{2}} cos^{2} (x) d x

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\int_{- B}^{B} - B^{2} - x^{2} d x

integral from 0 to 0.1 of 400x

\int_{0}^{0.1} 400 x d x

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integral from 0 to x of e^{-y}

\int_{0}^{x} e^{- y} d y