What is the integral from 0 to 2pi of-3sin(x) ?

The integral from 0 to 2pi of-3sin(x) is 0

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integral from 0 to 2pi of-3sin(x)

\int_{0}^{2 π} - 3 sin (x) d x

0

Solution steps

\int_{0}^{2 π} - 3 sin (x) d x

Take the constant out:

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

= - 3 \cdot \int_{0}^{2 π} sin (x) d x

Use the common integral:

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

= - 3 [- cos (x)]_{0}^{2 π}

Compute the boundaries:

0

= - 3 \cdot 0

Simplify

= 0

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\int_{1}^{5} (2 x + 1) ln (x + 1) d x

\int_{0}^{x} sin (t^{3}) d t

\int_{1}^{2} 8 - 2 x d x

\int_{\frac{π}{3}}^{π} sin (2 x) d x

What is the integral from 0 to 2pi of-3sin(x) ?
The integral from 0 to 2pi of-3sin(x) is 0