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人気のある微分積分 >

integral from 0 to 3 of sqrt((9-x^2))

解

\int_{0}^{3} (9 - x^{2}) d x

解

\frac{9 π}{4}

+1

十進法表記

7.06858 \dots

解答ステップ

\int_{0}^{3} 9 - x^{2} d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{2}} 9 cos^{2} (u) d u

定数を除く:

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

= 9 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 9 \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= 9 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

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

= 9 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= 9 \cdot \frac{1}{2} (\frac{π}{2} + 0)

簡素化

9 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{9 π}{4}

= \frac{9 π}{4}

グラフ

人気の例

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\int_{3}^{4} \frac{2}{x ^{2}} d x

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\int_{0}^{1} (5 x^{2} - 6 x + 6) d x

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\int_{- 1}^{1} x^{2} + 3 - 4 x^{2} d x

integral from 0 to 1 of cos(pinx)

\int_{0}^{1} cos (πn x) d x

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

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