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

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

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integral from 0 to pi of 2xsin(x)

\int_{0}^{π} 2 x sin (x) d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{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}^{π} x sin (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

= 2 [- x cos (x) + sin (x)]_{0}^{π}

Compute the boundaries:

π

= 2 π

\int_{1}^{10} ln (x) d x

\int_{- 1}^{1} ∣ x - 1 ∣ d x

\int_{0}^{2 π} x^{2} sin (x) d x

\int_{0}^{\frac{5 π}{6}} sin (x) d x

\int_{0}^{1} 8 cos (\frac{π x}{2}) d x

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