Let, y=limx→2π(x−2π)2∫x3(2π)3cost31dt
⇒y=limh→0(2π−h−2π)2∫(2π−h)3(2π)3cost31dt
⇒y=limh→0h2∫(2π−h)3(2π)3cost31dt
Applying L-Hospital's rule and Newton Leibnitz Theorem we get,
⇒y=limh→02h0+cos(2π−h)×3×(2π−h)2
⇒y=limh→02hsinh×3×(2π−h)2
We know that, limh→0(hsinh)=1
⇒y=limh→023×(2π−h)2
⇒y=83π2