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人気のある微分積分 >

integral from 0 to 6 of (16-2x)^{1/2}

解

\int_{0}^{6} (16 - 2 x)^{\frac{1}{2}} d x

解

\frac{56}{3}

+1

十進法表記

18.66666 \dots

解答ステップ

\int_{0}^{6} (16 - 2 x)^{\frac{1}{2}} d x

(16 - 2 x)^{\frac{1}{2}} = 2^{\frac{1}{2}} (8 - x)^{\frac{1}{2}}

= \int_{0}^{6} 2^{\frac{1}{2}} (8 - x)^{\frac{1}{2}} d x

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}^{6} (8 - x)^{\frac{1}{2}} d x

u置換積分法を適用する

= 2^{\frac{1}{2}} \cdot \int_{8}^{2} - u^{\frac{1}{2}} d u

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

= 2^{\frac{1}{2}} (- \int_{2}^{8} - u^{\frac{1}{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}} (- (- \int_{2}^{8} u^{\frac{1}{2}} d u))

べき乗則を適用する

= 2^{\frac{1}{2}} (- (- [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{8}))

簡素化

= 2^{\frac{1}{2}} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{8}

限度を計算する:

\frac{28 \cdot 2 ^{\frac{1}{2}}}{3}

= 2^{\frac{1}{2}} \cdot \frac{28 \cdot 2 ^{\frac{1}{2}}}{3}

簡素化

= \frac{56}{3}

グラフ

人気の例

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\int_{0}^{4} \frac{1}{96} x y d x

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