What is the integral from-i to i of (3z^2-1) ?

The integral from-i to i of (3z^2-1) is 2i^3-2i

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integral from-i to i of (3z^2-1)

\int_{- i}^{i} (3 z^{2} - 1) d z

2 i^{3} - 2 i

Solution steps

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

Apply the Sum Rule:

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

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

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

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

= 2 i^{3} - 2 i

\int_{1}^{9} \frac{6}{x} d x

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

\int_{- 3}^{0} 3 + 9 - x^{2} d x

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

\int_{- 1}^{1} x^{2} e^{x^{3}} d x

What is the integral from-i to i of (3z^2-1) ?
The integral from-i to i of (3z^2-1) is 2i^3-2i