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人気のある微分積分 >

integral from 0 to 1 of 2x(1-x^2)^4

解

\int_{0}^{1} 2 x (1 - x^{2})^{4} d x

解

\frac{1}{5}

+1

十進法表記

0.2

解答ステップ

\int_{0}^{1} 2 x (1 - x^{2})^{4} d x

定数を除く:

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

= 2 \cdot \int_{0}^{1} x (1 - x^{2})^{4} d x

u置換積分法を適用する

= 2 \cdot \int_{1}^{0} - \frac{u ^{4}}{2} d u

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

= 2 (- \int_{0}^{1} - \frac{u ^{4}}{2} d u)

定数を除く:

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

= 2 (- (- \frac{1}{2} \cdot \int_{0}^{1} u^{4} d u))

べき乗則を適用する

= 2 (- (- \frac{1}{2} [\frac{u ^{5}}{5}]_{0}^{1}))

簡素化

2 (- (- \frac{1}{2} [\frac{u ^{5}}{5}]_{0}^{1})) : [\frac{u ^{5}}{5}]_{0}^{1}

= [\frac{u ^{5}}{5}]_{0}^{1}

限度を計算する:

\frac{1}{5}

= \frac{1}{5}

グラフ

人気の例

integral from-1 to 2 of (3x^2-2x)

\int_{- 1}^{2} (3 x^{2} - 2 x) d x

integral from-1 to 1 of x^6

\int_{- 1}^{1} x^{6} d x

integral from 0 to 2 of |-x^2+3x-2|

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

integral from-2 to 2 of |x+1|

\int_{- 2}^{2} ∣ x + 1 ∣ d x

integral from 1 to 0 of |2x-1|

\int_{1}^{0} ∣ 2 x - 1 ∣ d x