Question

`int_0^{pi/2} xsinx dx` = ______

Options

• 1

• 2

• 3

• 4

Answer 1

`int_0^{pi/2} xsinx dx` = 1

Explanation:

`int_0^{pi/2} xsinx dx`

= `[x(-cosx)]_0^{pi/2} - int_0^{pi/2}1(-cosx)dx`

= `[-xcosx]_0^{pi/2} + [sinx]_0^{pi/2}`

= `[-pi/2cos(pi/2) + 0] + [sin  pi/2 - sin0]`

= 0 + 1 - 0 = 1