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人気のある微分積分 >

integral from 0 to 1 of sqrt(9-12x+4x^2)

解

\int_{0}^{1} 9 - 12 x + 4 x^{2} d x

解

2

解答ステップ

\int_{0}^{1} 9 - 12 x + 4 x^{2} d x

因数

9 - 12 x + 4 x^{2} : (2 x - 3)^{2}

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

指数の規則を適用する:

a^{2} = ∣ a ∣

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

絶対数を排除する

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

指数の規則を適用する:

a^{2} = ∣ a ∣

= \int_{0}^{1} ∣ (2 x - 3) ∣ d x

絶対数を排除する

= \int_{0}^{1} - 2 x + 3 d x

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

= - \int_{0}^{1} 2 x d x + \int_{0}^{1} 3 d x

\int_{0}^{1} 2 x d x = 1

\int_{0}^{1} 3 d x = 3

= - 1 + 3

簡素化

= 2

グラフ

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