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受欢迎的微积分 >

integral from 0 to 1 of sqrt(64t^2+9t^4)

解答

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

解答

\frac{73 73 - 512}{27}

+1

十进制

4.13749 \dots

求解步骤

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

分解

64 t^{2} + 9 t^{4} : t^{2} 64 + 9 t^{2}

= \int_{0}^{1} t^{2} 64 + 9 t^{2} d t

t^{2} = (t),

假定

t \geq 0

= \int_{0}^{1} t 64 + 9 t^{2} d t

使用换元积分法

= \int_{64}^{73} \frac{u}{18} d u

提出常数:

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

= \frac{1}{18} \cdot \int_{64}^{73} u d u

使用幂法则

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{64}^{73}

带入上下限计算后得:

\frac{146 73}{3} - \frac{1024}{3}

= \frac{1}{18} (\frac{146 73}{3} - \frac{1024}{3})

化简

= \frac{73 73 - 512}{27}

作图

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