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

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

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integral from 0 to 2pi of 2sin(t)

\int_{0}^{2 π} 2 sin (t) d t

0

Solution steps

\int_{0}^{2 π} 2 sin (t) d t

Take the constant out:

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

= 2 \cdot \int_{0}^{2 π} sin (t) d t

Use the common integral:

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

= 2 [- cos (t)]_{0}^{2 π}

Compute the boundaries:

0

= 2 \cdot 0

Simplify

= 0

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