Mate 2
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2) Para la función f(x) = X2 con [0,4] calcular la suma de Reimann para dicha función en las siguientes partituras: x x =2;x =2,5; x = 3 4 5 4 ; y para las t1=0.5
f(x)= x2
0=
0; x =0,5; x =1,5; 1 2
;t2=1,5;t3=1,75;t4=2,5.
; [0, 4]
i=14fti∆ix=f0.50.5-0+f1.51.5-0.5+f1.752-1.5+f2.52.5-2 i=14fti∆x=f0.50.5+f1.51+f1.750.5+f2.5(0.5)
F(x)= x2
F(t)= t2
F(0)= 02 =0
F(0.5)= 0.52=0.25
F(0.5)= 0.52= 0.25
F(1.5)= 1.52=2.25
F(1.5)= 1.52=2.25
F(1.75)= 1.752=3.06
F(2)= 22=4
F(2.5)= 2.52=6.25
F(2.5)= 2.52=6.25
i=14fti∆x=f0.250.5+f2.251+f3.060.5+f6.25(0.5) i=14fti∆x=f18+f94+f1.53+f3.125 i=14fti∆x=7.03
f(x)= x2
0=
0; x =0,5; x =1,5; 1 2
;t2=1,5;t3=1,75;t4=2,5.
; [0, 4]
i=14fti∆ix=f0.50.5-0+f1.51.5-0.5+f1.752-1.5+f2.52.5-2 i=14fti∆x=f0.50.5+f1.51+f1.750.5+f2.5(0.5)
F(x)= x2
F(t)= t2
F(0)= 02 =0
F(0.5)= 0.52=0.25
F(0.5)= 0.52= 0.25
F(1.5)= 1.52=2.25
F(1.5)= 1.52=2.25
F(1.75)= 1.752=3.06
F(2)= 22=4
F(2.5)= 2.52=6.25
F(2.5)= 2.52=6.25
i=14fti∆x=f0.250.5+f2.251+f3.060.5+f6.25(0.5) i=14fti∆x=f18+f94+f1.53+f3.125 i=14fti∆x=7.03
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