Q.

The integral ∫–13/2(|π2xsin(πx)|)dx is equal to :          [2025]

1 3+2π  
2 4+π  
3 1+3π  
4 2+3π  

Ans.

(3)

Let I=∫–13/2|π2xsinπx|dx

=π2[∫–11xsinπxdx–∫13/2xsinπxdx]

=π2[2∫01xsinπxdx–∫13/2xsinπxdx]

=π2[2[–xcosπxπ+sinπxπ2]01–[–xcosπxπ+sinπxπ2]13/2]

=π2[2π–(–1π2–1π)]=π2[3π+1π2]

=3π+1.