∫ 0 π si n 2 (x) ⁢ ⅆ x = ∫ 0 π 1-cos( 2x ) 2 ⁢ ⅆ x = 1 2 [ x- sin( 2x ) 2 ] 0 π = π 2


Ampiezza del sottointervallo: ∆x= (b - a)/n= π/10
  • Metodo dei rettangoli

∫ 0 π si n 2 (x) ⁢ ⅆ x = π 10 [ f⁡(0)+f⁡( π 10 )+f⁡( 2 π 10 )+f⁡( 3 π 10 )+f⁡( 4 π 10 )+f⁡( 5 π 10 )+f⁡( 6 π 10 )+f⁡( 7 π 10 )+f⁡( 8 π 10 )+f⁡( 9 π 10 ) ]=

= π 10 [ si n 2 (0)+si n 2 ( π 10 )+si n 2 ( 2 π 10 )+si n 2 ( 3 π 10 )+si n 2 ( 4 π 10 )+si n 2 ( 5 π 10 )+si n 2 ( 6 π 10 )+si n 2 ( 7 π 10 )+si n 2 ( 8 π 10 )+si n 2 ( 9 π 10 ) ]=1.5707963268


  • Metodo dei trapezi

∫ 0 π si n 2 (x) ⁢ ⅆ x = π 10 [ f⁡(0)+f⁡( π ) 2 +f⁡( π 10 )+f⁡( 2 π 10 )+f⁡( 3 π 10 )+f⁡( 4 π 10 )+f⁡( 5 π 10 )+f⁡( 6 π 10 )+f⁡( 7 π 10 )+f⁡( 8 π 10 )+f⁡( 9 π 10 ) ]=
= π 10 [ si n 2 (0)+si n 2 ( π ) 2 +si n 2 ( π 10 )+si n 2 ( 2 π 10 )+si n 2 ( 3 π 10 )+si n 2 ( 4 π 10 )+si n 2 ( 5 π 10 )+si n 2 ( 6 π 10 )+si n 2 ( 7 π 10 )+si n 2 ( 8 π 10 )+si n 2 ( 9 π 10 ) ]=1.5707963268


  • Metodo delle parabole
∫ 0 π si n 2 (x) ⁢ ⅆ x = 1 3 ⋅ π 10 { f⁡(0)+f⁡( π )+2 ⋅ [ f⁡( 2 π 10 )+f⁡( 4 π 10 )+f⁡( 6 π 10 )+f⁡( 8 π 10 ) ]+4 ⋅ [ f⁡( π 10 )+f⁡( 3 π 10 )+f⁡( 5 π 10 )+f⁡( 7 π 10 )+f⁡( 9 π 10 ) ] }=
= 1 3 ⋅ π 10 { si n 2 (0)+si n 2 ( π )+2 ⋅ [ si n 2 ( 2 π 10 )+si n 2 ( 4 π 10 )+si n 2 ( 6 π 10 )+si n 2 ( 8 π 10 ) ]+4 ⋅ [ si n 2 ( π 10 )+si n 2 ( 3 π 10 )+si n 2 ( 5 π 10 )+si n 2 ( 7 π 10 )+si n 2 ( 9 π 10 ) ] }=1.5707963268