0 1 1-x 1+x x = π 2 -1 π =2( 1+ 0 1 1-x 1+x x )


Ampiezza del sottointervallo: ∆x= (b - a)/n= 1/10= 0.1


  • Metodo dei rettangoli

0 1 1-x 1+x x = 1 10 [ f(0)+f(0.1)+f(0.2)+f(0.3)+f(0.4)+f(0.5)+f(0.6)+f(0.7)+f(0.8)+f(0.9) ]=

= 1 10 [ 1-0 1+0 + 1-0.1 1+0.1 + 1-0.2 1+0.2 + 1-0.3 1+0.3 + 1-0.4 1+0.4 + 1-0.5 1+0.5 + 1-0.6 1+0.6 + 1-0.7 1+0.7 + 1-0.8 1+0.8 + 1-0.9 1+0.9 ]0.6169667033
π =2( 1+ 0 1 1-x 1+x x )3.2339334065


  • Metodo dei trapezi

0 1 1-x 1+x x = 1 10 [ f(0)+f(1) 2 +f(0.1)+f(0.2)+f(0.3)+f(0.4)+f(0.5)+f(0.6)+f(0.7)+f(0.8)+f(0.9) ]=
= 1 10 [ 1-0 1+0 + 1-1 1+1 2 + 1-0.1 1+0.1 + 1-0.2 1+0.2 + 1-0.3 1+0.3 + 1-0.4 1+0.4 + 1-0.5 1+0.5 + 1-0.6 1+0.6 + 1-0.7 1+0.7 + 1-0.8 1+0.8 + 1-0.9 1+0.9 ]0.5669667033
π =2( 1+ 0 1 1-x 1+x x )3.1339334065


  • Metodo delle parabole
0 1 1-x 1+x x = 1 30 { f(0)+f(1)+2[ f(0.2)+f(0.4)+f(0.6)+f(0.8) ]+4[ f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9) ] }=
= 1 10 { 1-0 1+0 + 1-1 1+1 +2[ 1-0.2 1+0.2 + 1-0.4 1+0.4 + 1-0.6 1+0.6 + 1-0.8 1+0.8 ]+4[ 1-0.1 1+0.1 + 1-0.3 1+0.3 + 1-0.5 1+0.5 + 1-0.7 1+0.7 + 1-0.9 1+0.9 ] }0.568990032
π =2( 1+ 0 1 1-x 1+x x )3.1379800641