What is the integral from 0 to 2pi of ((9θ)/2+81/4) ?

The integral from 0 to 2pi of ((9θ)/2+81/4) is 9pi^2+(81pi)/2

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integral from 0 to 2pi of ((9θ)/2+81/4)

\int_{0}^{2 π} (\frac{9 θ}{2} + \frac{81}{4}) d θ

9 π^{2} + \frac{81 π}{2}

Decimal

216.06094 \dots

Solution steps

\int_{0}^{2 π} \frac{9 θ}{2} + \frac{81}{4} d θ

Apply the Sum Rule:

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

= \int_{0}^{2 π} \frac{9 θ}{2} d θ + \int_{0}^{2 π} \frac{81}{4} d θ

\int_{0}^{2 π} \frac{9 θ}{2} d θ = 9 π^{2}

\int_{0}^{2 π} \frac{81}{4} d θ = \frac{81 π}{2}

= 9 π^{2} + \frac{81 π}{2}

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What is the integral from 0 to 2pi of ((9θ)/2+81/4) ?
The integral from 0 to 2pi of ((9θ)/2+81/4) is 9pi^2+(81pi)/2