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受欢迎的微积分 >

integral from 0 to 5 of sqrt(9t^4+16t^2)

解答

\int_{0}^{5} 9 t^{4} + 16 t^{2} d t

解答

\frac{241 241 - 64}{27}

+1

十进制

136.19726 \dots

求解步骤

\int_{0}^{5} 9 t^{4} + 16 t^{2} d t

分解

9 t^{4} + 16 t^{2} : t^{2} 9 t^{2} + 16

= \int_{0}^{5} t^{2} 9 t^{2} + 16 d t

t^{2} = (t),

假定

t \geq 0

= \int_{0}^{5} t 9 t^{2} + 16 d t

使用换元积分法

= \int_{16}^{241} \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_{16}^{241} u d u

使用幂法则

= \frac{1}{18} [\frac{2}{3} u^{\frac{3}{2}}]_{16}^{241}

带入上下限计算后得:

\frac{482 241}{3} - \frac{128}{3}

= \frac{1}{18} (\frac{482 241}{3} - \frac{128}{3})

化简

= \frac{241 241 - 64}{27}

作图

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