integral from 0 to 1+i of z^2

解答

\int_{0}^{1 + i} z^{2} d z

\frac{( i + 1 ) ^{3}}{3}

求解步骤

\int_{0}^{1 + i} z^{2} d z

使用幂法则

= [\frac{z ^{3}}{3}]_{0}^{1 + i}

带入上下限计算后得:

\frac{( i + 1 ) ^{3}}{3}

= \frac{( i + 1 ) ^{3}}{3}

\int_{- 1}^{1} e^{x} \cdot x d x

\int_{0}^{1} (3 x - 3 x^{2}) d x

\int_{0}^{π} (π^{2} - x^{2}) d x

\int_{- π}^{π} cos (2 x) cos (2 x) d x

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