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

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

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

\int_{0}^{2 π} (x sin (x)) d x

- 2 π

Decimal Notation

- 6.28318 \dots

Solution steps

\int_{0}^{2 π} x sin (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

= [- x cos (x) + sin (x)]_{0}^{2 π}

Compute the boundaries:

- 2 π

= - 2 π

\int_{0}^{t a n (x)} \frac{d t}{1 + t ^{2}} d x

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

\int_{0}^{1} - 4 x d x

\int_{0}^{2 y} 1 d x

\int_{2}^{B} (2 x - 3)^{3} d x

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