The value of ∑r=120(|π(∫0rx|sinπx|dx)|) is _______. [2026]
(210)
Let Ir=∫0rx|sinπx|dx ...(1)
Apply King Property
=∫0r(r-x)|sinπx|dx ...(2)
By (1) + (2)
2Ir=∫0rr|sinπx|dx ⇒ Ir=r2∫0r|sinπx|dx
I1=12∫01|sinπx|dx =12π∫0π|sint|dt=12π(2)
I2=22∫02|sinπx|dx =22π∫02π|sint| dt=22π(4)
S=π·12π·2+π·22π·4+π·32π·6+⋯+π·202π·(2·20)
=1+2+3+⋯+20
=20×212=210