integral from 1 to 5 of sqrt(9-6x+x^2)

解答

\int_{1}^{5} 9 - 6 x + x^{2} d x

4

求解步骤

\int_{1}^{5} 9 - 6 x + x^{2} d x

分解

9 - 6 x + x^{2} : (x - 3)^{2}

= \int_{1}^{5} (x - 3)^{2} d x

使用指数法则:

a^{2} = ∣ a ∣

= \int_{1}^{5} (x - 3)^{2} d x

拿掉绝对值

= \int_{1}^{3} (x - 3)^{2} d x + \int_{3}^{5} (x - 3)^{2} d x

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

\int_{3}^{5} (x - 3)^{2} d x = 2

= 2 + 2

化简

= 4

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