Anonymous
Anonymous asked in Science & MathematicsOther - Science · 2 months ago

Calculus help.....?

Please help me !

Thanks in advance !

Attachment image

1 Answer

Relevance
  • Indica
    Lv 7
    2 months ago
    Best 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

    • S.K.D2 months agoReport

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

Still have questions? Get your answers by asking now.