integral from 1 to 2 of t^2sqrt(t^3+1)

解答

\int_{1}^{2} t^{2} t^{3} + 1 d t

\frac{54 - 4 2}{9}

十进制

5.37146 \dots

求解步骤

\int_{1}^{2} t^{2} t^{3} + 1 d t

使用换元积分法

= \int_{2}^{9} \frac{u}{3} d u

提出常数:

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

= \frac{1}{3} \cdot \int_{2}^{9} u d u

使用幂法则

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{2}^{9}

带入上下限计算后得:

18 - \frac{4 2}{3}

= \frac{1}{3} (18 - \frac{4 2}{3})

化简

= \frac{54 - 4 2}{9}

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