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

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

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

\int_{0}^{π} 2 x sin (2 x) d x

- π

Decimal

- 3.14159 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{π}

= 2 [\frac{1}{2} (- x cos (2 x) - 2 \cdot \int - \frac{1}{2} cos (2 x) d x)]_{0}^{π}

\int - \frac{1}{2} cos (2 x) d x = - \frac{1}{4} sin (2 x)

= 2 [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{π}

Simplify

2 [\frac{1}{2} (- x cos (2 x) - 2 (- \frac{1}{4} sin (2 x)))]_{0}^{π} : 2 [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{π}

= 2 [\frac{1}{2} (- x cos (2 x) + \frac{1}{2} sin (2 x))]_{0}^{π}

Compute the boundaries:

- \frac{π}{2}

= 2 (- \frac{π}{2})

Simplify

= - π

\int_{0}^{5} 1 - - x^{2} + 5 x - 4 d x

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\int_{2}^{e} \frac{x ^{4} - 5 x}{x ^{2}} d x

\int_{0}^{1} x^{- \frac{1}{2}} ln (x) d x

\int_{- \frac{π}{3}}^{\frac{π}{3}} sin (x) d x

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