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# Calculus help.....?

Please help me !

Thanks in advance !

### 1 Answer

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- 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

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I got sir ! Infinite thanks dear sir ! :))