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受欢迎的微积分 >

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

解答

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

解答

\frac{5 5 - 1}{6}

+1

十进制

1.69672 \dots

求解步骤

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

提出常数:

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

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

使用换元积分法

= 2 \cdot \int_{1}^{5} \frac{u}{8} d u

提出常数:

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

= 2 \cdot \frac{1}{8} \cdot \int_{1}^{5} u d u

使用幂法则

= 2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

化简

2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5} : \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

带入上下限计算后得:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{10 5}{3} - \frac{2}{3})

化简

= \frac{5 5 - 1}{6}

作图

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