integral from-1 to 2 of x^3sqrt(x^4+3)

解答

\int_{- 1}^{2} x^{3} x^{4} + 3 d x

\frac{19 19 - 8}{6}

十进制

12.46984 \dots

求解步骤

\int_{- 1}^{2} x^{3} x^{4} + 3 d x

使用换元积分法

= \int_{4}^{19} \frac{u}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{4}^{19} u d u

使用幂法则

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{19}

带入上下限计算后得:

\frac{38 19}{3} - \frac{16}{3}

= \frac{1}{4} (\frac{38 19}{3} - \frac{16}{3})

化简

= \frac{19 19 - 8}{6}

\int_{- 4}^{- 3} \frac{1}{x - 3} d x

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