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人気のある微分積分 >

integral from 0 to 4 of sqrt(2-\sqrt{x)}

解

\int_{0}^{4} 2 - x d x

解

\frac{32 2}{15}

+1

十進法表記

3.01698 \dots

解答ステップ

\int_{0}^{4} 2 - x d x

u置換積分法を適用する

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

定数を除く:

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

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

u置換積分法を適用する

= 2 \cdot \int_{2}^{0} - (- v + 2) v d v

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

= 2 (- \int_{0}^{2} - (- v + 2) v d v)

定数を除く:

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

= 2 (- (- \int_{0}^{2} (- v + 2) v d v))

拡張

(- v + 2) v : - v^{\frac{3}{2}} + 2 v

= 2 (- (- \int_{0}^{2} - v^{\frac{3}{2}} + 2 v d v))

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- (- (- \int_{0}^{2} v^{\frac{3}{2}} d v + \int_{0}^{2} 2 v d v)))

\int_{0}^{2} v^{\frac{3}{2}} d v = \frac{8 2}{5}

\int_{0}^{2} 2 v d v = \frac{8 2}{3}

= 2 (- (- (- \frac{8 2}{5} + \frac{8 2}{3})))

簡素化

2 (- (- (- \frac{8 2}{5} + \frac{8 2}{3}))) : \frac{32 2}{15}

= \frac{32 2}{15}

グラフ

人気の例

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\int_{1}^{5} 3^{2 x} d x

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