What is the integral from 0 to 2pi of ((pi-x)/2) ?

The integral from 0 to 2pi of ((pi-x)/2) is 0

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integral from 0 to 2pi of ((pi-x)/2)

\int_{0}^{2 π} (\frac{π - x}{2}) d x

0

Solution steps

\int_{0}^{2 π} \frac{π - x}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2 π} π - x d x

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{2 π} π d x - \int_{0}^{2 π} x d x)

\int_{0}^{2 π} π d x = 2 π^{2}

\int_{0}^{2 π} x d x = 2 π^{2}

= \frac{1}{2} (2 π^{2} - 2 π^{2})

Simplify

= 0

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What is the integral from 0 to 2pi of ((pi-x)/2) ?
The integral from 0 to 2pi of ((pi-x)/2) is 0