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

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

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

\int_{0}^{2 π} - sin (x) d x

0

Solution steps

\int_{0}^{2 π} - sin (x) d x

Take the constant out:

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

= - \int_{0}^{2 π} sin (x) d x

Use the common integral:

\int sin (x) d x = - cos (x)

= - [- cos (x)]_{0}^{2 π}

Compute the boundaries:

0

= - 0

Simplify

= 0

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