What is the integral from 2pi to-4pi of sin(x) ?

The integral from 2pi to-4pi of sin(x) is 0

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integral from 2pi to-4pi of sin(x)

\int_{2 π}^{- 4 π} sin (x) d x

0

Solution steps

\int_{2 π}^{- 4 π} sin (x) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 4 π}^{2 π} sin (x) d x

Use the common integral:

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

= - [- cos (x)]_{- 4 π}^{2 π}

Compute the boundaries:

0

= - 0

Simplify

= 0

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\int_{0}^{3} 2 x x^{2} + 16 d x

\int_{4}^{5} x (x^{2} - 16)^{3} d x

What is the integral from 2pi to-4pi of sin(x) ?
The integral from 2pi to-4pi of sin(x) is 0