integral from i to 1 of (z+1)^2

解

\int_{i}^{1} (z + 1)^{2} d z

- i^{2} - i + \frac{- i ^{3} + 1}{3} + 2

解答ステップ

\int_{i}^{1} (z + 1)^{2} d z

拡張

(z + 1)^{2} : z^{2} + 2 z + 1

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

総和規則を適用する:

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

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

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

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

\int_{i}^{1} 1 d z = 1 - i

= \frac{1 - i ^{3}}{3} + 1 - i^{2} + 1 - i

簡素化

= - i^{2} - i + \frac{- i ^{3} + 1}{3} + 2