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人気のある微分積分 >

integral from 0 to 1 of (2-x^2)^{3/2}

解

\int_{0}^{1} (2 - x^{2})^{\frac{3}{2}} d x

解

\frac{6 π + 16}{16}

+1

十進法表記

2.17809 \dots

解答ステップ

\int_{0}^{1} (2 - x^{2})^{\frac{3}{2}} d x

置換積分法を適用する

= \int_{0}^{2} \frac{( 8 - u ^{2} ) ^{\frac{3}{2}}}{16} d u

定数を除く:

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

= \frac{1}{16} \cdot \int_{0}^{2} (8 - u^{2})^{\frac{3}{2}} d u

部分積分を適用する

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u]_{0}^{2}

\int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u = - 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - (- 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}))))]_{0}^{2}

簡素化

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} + 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))]_{0}^{2}

限度を計算する:

6 π + 16

= \frac{1}{16} (6 π + 16)

簡素化

= \frac{6 π + 16}{16}

グラフ

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