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

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

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integral from 0 to 2pi of 2xsin(x)

\int_{0}^{2 π} 2 x sin (x) d x

- 4 π

Decimal

- 12.56637 \dots

Solution steps

\int_{0}^{2 π} 2 x sin (x) d x

Take the constant out:

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

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

Apply Integration By Parts

: [- x cos (x) - \int - cos (x) d x]_{0}^{2 π}

= 2 [- x cos (x) - \int - cos (x) d x]_{0}^{2 π}

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

= 2 [- x cos (x) - (- sin (x))]_{0}^{2 π}

Simplify

= 2 [- x cos (x) + sin (x)]_{0}^{2 π}

Compute the boundaries:

- 2 π

= 2 (- 2 π)

Simplify

= - 4 π

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