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

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

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

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

2

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

1

= 2 \cdot 1

Simplify

= 2

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