Please help me !
Thanks in advance !
- IndicaLv 72 months agoBest Answer
I = ∫ [x=0,π] |sin(x)/x| dx
Divide [0,π] into n intervals of size π/n
I = ∑ [r=1,n] ∫ [x=(r−1)π/n,rπ/n] |sin(nx)/x| dx
Apply the sub u=nx and this becomes I = ∑ [r=1,n] ∫ [x=(r−1)π,rπ] |sin(u)/u| du
(note that the integral for r=1 exists since sin(x)/x→1 as x→0)
But ∫ [x=(r−1)π,rπ] |sin(u)/u| du ≥ (1/rπ) ∫ [x=(r−1)π,rπ] |sin(u)| du = 2/(rπ)
∴ I ≥ ∑ [r=1,n] (2/rπ) which is desired result