What is the integral from 0 to 2pi of-9 ?

The integral from 0 to 2pi of-9 is -18pi

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integral from 0 to 2pi of-9

\int_{0}^{2 π} - 9 d t

- 18 π

Solution steps

\int_{0}^{2 π} - 9 d t

Integral of a constant:

\int a d x = a x

= [(- 9) t]_{0}^{2 π}

Simplify

= [- 9 t]_{0}^{2 π}

Compute the boundaries:

- 18 π

= - 18 π

\int_{1}^{\infty} t e^{- s t} d t

\int_{4}^{10} ∣ sin (x) ∣ d x

\int_{1}^{5} 4 d x

\int_{1}^{e} \frac{ln ( x )}{2 x} d x

\int_{2}^{4} (3 x - 1 + 2 x^{2}) d x

What is the integral from 0 to 2pi of-9 ?
The integral from 0 to 2pi of-9 is -18pi