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人気のある微分積分 >

integral from-2 to 2 of (sqrt(8-x^2))

解

\int_{- 2}^{2} (8 - x^{2}) d x

解

4 (\frac{π}{2} + 1)

+1

十進法表記

10.28318 \dots

解答ステップ

\int_{- 2}^{2} 8 - x^{2} d x

三角関数による置換を適用する

= \int_{- \frac{π}{4}}^{\frac{π}{4}} 8 cos^{2} (u) d u

定数を除く:

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

= 8 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 8 \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} \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

= 8 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{\frac{π}{4}} 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

= 8 \cdot \frac{1}{2} (\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d u + \int_{- \frac{π}{4}}^{\frac{π}{4}} cos (2 u) d u)

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

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

= 8 \cdot \frac{1}{2} (\frac{π}{2} + 1)

簡素化

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

= 4 (\frac{π}{2} + 1)

グラフ

人気の例

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\int_{1}^{\infty} \frac{1}{6 ^{x}} d x

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\int_{\frac{- π}{2}}^{\frac{π}{2}} 1 d x

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