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

The integral from-2 to i/2 of (z^2+1) is (i^3)/(24)+8/3+i/2+2

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

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

\frac{i ^{3}}{24} + \frac{8}{3} + \frac{i}{2} + 2

Solution steps

\int_{- 2}^{\frac{i}{2}} 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_{- 2}^{\frac{i}{2}} z^{2} d z + \int_{- 2}^{\frac{i}{2}} 1 d z

\int_{- 2}^{\frac{i}{2}} z^{2} d z = \frac{i ^{3}}{24} + \frac{8}{3}

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

= \frac{i ^{3}}{24} + \frac{8}{3} + \frac{i}{2} + 2

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What is the integral from-2 to i/2 of (z^2+1) ?
The integral from-2 to i/2 of (z^2+1) is (i^3)/(24)+8/3+i/2+2