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受欢迎的微积分 >

integral from 0 to 6 of sqrt(1+x^4+2x^2)

解答

\int_{0}^{6} 1 + x^{4} + 2 x^{2} d x

解答

78

求解步骤

\int_{0}^{6} 1 + x^{4} + 2 x^{2} d x

分解

1 + x^{4} + 2 x^{2} : (x^{2} + 1)^{2}

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

使用指数法则:

a^{2} = ∣ a ∣

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

拿掉绝对值

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

使用指数法则:

a^{2} = ∣ a ∣

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

拿掉绝对值

= \int_{0}^{6} x^{2} + 1 d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{6} x^{2} d x + \int_{0}^{6} 1 d x

\int_{0}^{6} x^{2} d x = 72

\int_{0}^{6} 1 d x = 6

= 72 + 6

化简

= 78

作图

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