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受欢迎的微积分 >

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

解答

\int_{0}^{1} 4 + 4 t^{2} + t^{4} d t

解答

\frac{7}{3}

+1

十进制

2.33333 \dots

求解步骤

\int_{0}^{1} 4 + 4 t^{2} + t^{4} d t

分解

4 + 4 t^{2} + t^{4} : (t^{2} + 2)^{2}

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

使用指数法则:

a^{2} = ∣ a ∣

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

拿掉绝对值

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

使用指数法则:

a^{2} = ∣ a ∣

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

拿掉绝对值

= \int_{0}^{1} t^{2} + 2 d t

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}^{1} t^{2} d t + \int_{0}^{1} 2 d t

\int_{0}^{1} t^{2} d t = \frac{1}{3}

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

= \frac{1}{3} + 2

化简

\frac{1}{3} + 2 : \frac{7}{3}

= \frac{7}{3}

作图

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