∫ 1 e l n 2 (x) ⁢ ⅆ x { F⁡(x)=l n 2 (x) g⁡(x)=1 f⁡(x)= 2 ⋅ ln⁡(x) x G⁡(x)=x ⇒ ∫ 1 e l n 2 (x) ⁢ ⅆ x = [ x ⋅ l n 2 (x) ] 1 e - ∫ 1 e ln⁡(x) ⁢ ⅆ x
 
∫ 1 e ln⁡(x) ⁢ ⅆ x { F⁡(x)=ln⁡(x) g⁡(x)=1 f⁡(x)= 1 x G⁡(x)=x ⇒ ∫ 1 e ln⁡(x) ⁢ ⅆ x = [ x ⋅ ln⁡(x) ] 1 e - ∫ 1 e ⁢ ⅆ x = [ x ⋅ ln⁡(x)-x ] 1 e
∫ 1 e l n 2 (x) ⁢ ⅆ x = [ x ⋅ l n 2 (x) ] 1 e -2 [ x ⋅ ln⁡(x)-x ] 1 e =e ⋅ l n 2 (e)-2e ⋅ ln⁡(e)+2e-2=e-2 ≃ 0.71828182846



Ampiezza del sottointervallo: ∆x= (b - a)/n= (e - 1)/10
  • Metodo dei rettangoli

∫ 1 e l n 2 (x) ⁢ ⅆ x = e-1 10 { f⁡(1)+f( 1+ e-1 10 )+f( 1+2 ⋅ e-1 10 )+f( 1+3 ⋅ e-1 10 )+f( 1+4 ⋅ e-1 10 )+f( 1+5 ⋅ e-1 10 )+f( 1+6 ⋅ e-1 10 )+f( 1+7 ⋅ e-1 10 )+f( 1+8 ⋅ e-1 10 )+f( 1+9 ⋅ e-1 10 ) }=


= e-1 10 { l n 2 (1)+l n 2 ( 1+ e-1 10 )+l n 2 ( 1+2 ⋅ e-1 10 )+l n 2 ( 1+3 ⋅ e-1 10 )+l n 2 ( 1+4 ⋅ e-1 10 )+l n 2 ( 1+5 ⋅ e-1 10 )+l n 2 ( 1+6 ⋅ e-1 10 )+l n 2 ( 1+7 ⋅ e-1 10 )+l n 2 ( 1+8 ⋅ e-1 10 )+l n 2 ( 1+9 ⋅ e-1 10 ) }=0.6341709447


  • Metodo dei trapezi

∫ 1 e l n 2 (x) ⁢ ⅆ x = e-1 10 { f⁡(1)+f⁡(e) 2 +f( 1+ e-1 10 )+f( 1+2 ⋅ e-1 10 )+f( 1+3 ⋅ e-1 10 )+f( 1+4 ⋅ e-1 10 )+f( 1+5 ⋅ e-1 10 )+f( 1+6 ⋅ e-1 10 )+f( 1+7 ⋅ e-1 10 )+f( 1+8 ⋅ e-1 10 )+f( 1+9 ⋅ e-1 10 ) }=
= e-1 10 { l n 2 (1)+l n 2 (e) 2 +l n 2 ( 1+ e-1 10 )+l n 2 ( 1+2 ⋅ e-1 10 )+l n 2 ( 1+3 ⋅ e-1 10 )+l n 2 ( 1+4 ⋅ e-1 10 )+l n 2 ( 1+5 ⋅ e-1 10 )+l n 2 ( 1+6 ⋅ e-1 10 )+l n 2 ( 1+7 ⋅ e-1 10 )+l n 2 ( 1+8 ⋅ e-1 10 )+l n 2 ( 1+9 ⋅ e-1 10 ) }=0.7200850361


  • Metodo delle parabole
∫ 1 e l n 2 (x) ⁢ ⅆ x = e-1 30 { f⁡(1)+f⁡(e)+2 ⋅ [ f( 1+2 ⋅ e-1 10 )+f( 1+4 ⋅ e-1 10 )+f( 1+6 ⋅ e-1 10 )+f( 1+8 ⋅ e-1 10 ) ]+4 ⋅ [ f( 1+ e-1 10 )+f( 1+3 ⋅ e-1 10 )+f( 1+5 ⋅ e-1 10 )+f( 1+7 ⋅ e-1 10 )+f( 1+9 ⋅ e-1 10 ) ] }=
= e-1 30 { l n 2 (1)+l n 2 (e)+2 ⋅ [ l n 2 ( 1+2 ⋅ e-1 10 )+l n 2 ( 1+4 ⋅ e-1 10 )+l n 2 ( 1+6 ⋅ e-1 10 )+l n 2 ( 1+8 ⋅ e-1 10 ) ]+4 ⋅ [ l n 2 ( 1+ e-1 10 )+l n 2 ( 1+3 ⋅ e-1 10 )+l n 2 ( 1+5 ⋅ e-1 10 )+l n 2 ( 1+7 ⋅ e-1 10 )+l n 2 ( 1+9 ⋅ e-1 10 ) ] }=0.7183088659