integral from 1-i to 2i of z^2-6-4iz

解答

\int_{1 - i}^{2 i} z^{2} - 6 - 4 i z d z

- 12 i + \frac{32}{3}

求解步骤

\int_{1 - i}^{2 i} z^{2} - 6 - 4 i z d z

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1 - i}^{2 i} z^{2} d z - \int_{1 - i}^{2 i} 6 d z - \int_{1 - i}^{2 i} 4 i z d z

\int_{1 - i}^{2 i} z^{2} d z = \frac{2}{3} - 2 i

\int_{1 - i}^{2 i} 6 d z = 18 i - 6

\int_{1 - i}^{2 i} 4 i z d z = - 4 - 8 i

= \frac{2}{3} - 2 i - (18 i - 6) - (- 4 - 8 i)

化简

\frac{2}{3} - 2 i - (18 i - 6) - (- 4 - 8 i) : - 12 i + \frac{32}{3}

= - 12 i + \frac{32}{3}

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