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NAMA:NANDA SASTAMA KELAS:2EGD NIM:061640411931 DOSEN:DR. YULIANTO.W MATEMATIKA NUMERIK
Selesaikan integral berikut dengan metode trafezoid dan metode simpson dengan n : 4 5
1. ∫2 𝑥 2 (√𝑥 − 2 ) 𝑏−𝑎 𝑛
h=
= 5-2/4 = 0,75
f(2) = 22 ( √2 − 2) = 0 f(2,75) = (2,75^3)*(2,75-2)^0,5 = 3,949386819 f(3,5) = (3,5^3)*(3,5-2)^0,5 =8,019507466 f(4,25) = (4,25^3)*(4,25-2)^0,5 =13,14239918 f(5) = (5^3)*(5-2)^0,5 = 19,36491673 X 2 2,75 3,5 4,25 5 -
f(x) = (x^3)*(x_2)^0,5 0 3,949386819 8,019507466 13,14239918 19,36491673
METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 0,3
L=(( 2 )) x(0+2*(3,949386819+8,019507466+13,14239918)+ 19,36491673) = 26,09531 -
METODE SIMPSON
-
L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn)
L = (0,3/3)*( 0+4*(3,949386819+13,14239918)+2*(8,019507466)+ 19,36491673) = 25,94277
2. f(x) = 3
∫ x^3 ∗ ln(2x + 1) 1
PENYELESAIAN: h = (b-a)/n = (3-1)/4 = 0,5 f(1) = (1^3)*(log(2*1+1)) = 0,477121255 f(1,5) = (1,5^3)*(log(2*1,5+1)) =2,031952471 f(2) = (2^3)*(log(2*2+1)) =5,591760035 f(2,5) = (2,5^3)*(log(2*2,5+1)) =12,15861329 f(3) = (3^3)*(log(2*3+1)) =22,81764708 X
f(x) = x^3*log(2x+1) 0,477121255 2,031952471 5,591760035 12,15861329 22,81764708
1 1,5 2 2,5 3 - METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) L=(0,2/2)(x(0,477121255+2*(2,031952471+5,591760035+12,15861329))+ 22,81764708) = 15,71485 METODE SIMPSON L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn) L = (0,2/3)*( 2,031952471+12,15861329)+2*(5,591760035+)+22,81764708) = 15,20676
3. f(x) = 5
∫(3^2x + 1)/(3x + 2) 3
PENYELESAIAN : h = (b-a)/10 = (5-3)/4 = 0,5 f(3) = (3^2*(3)+1)/(3*(3)+2) = 2,545454545 f(3,5) = (3^2*(3,5)+1)/(3*(3,5)+2) = 2,6 f(4) = (3^2*(4)+1)/(3*(4)+2) =2,642857143 f(4,5) = (3^2*(4,5)+1)/(3*(4,5)+2) =2,677419355 f(5) = (3^2*(5)+1)/(3*(5)+2) = 2,705882353 X 3 3,5 4 4,5 5 -
f(x) = (3^2x+1)/(3x+2) 2,545454545 2,6 2,642857143 2,677419355 2,705882353 METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) L=(0,2/2)(x(2,545454545+2*(2,6+2,642857143+2,677419355+)+2,705882353) = 5,272972 -
METODE SIMPSON L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn)
L = (0,2/3)*( 2,545454545+4*(2,6+2,677419355)+2*(2,642857143)+( 2,705882353) = 6,155407
Selesaikan integral berikut dengan metode trafezoid dan metode simpson dengan n : 10 4. f(x)=(x+1)^1,5/(2x+1)^0,5 Metode trapezoid dalam metode Trapezoid dan Simpson 1/3 ; n =10 metode trapezoid x 1 1,3 1,6 1,9 2,2 2,5 2,8 3,1 3,4 3,7 4 L 𝑏−𝑎
h=
10
=
f(x)=(x+1)^1,5/(2x+1)^0,5 1,632993162 1,838402084 2,04566906 2,254116604 2,463361149 2,673169155 2,88339026 3,093923256 3,304697511 3,515662235 3,726779962 8,025683363 4−1 10
=0,3
f(1)=(1+1)1,5/(2*1+1)0,5=1,632993162 f(1,3)=(1,3+1)1,5/(2*1,3+1)0,5=1,838402084 f(1,6)=(1,6+1)1,5/(2*1,6+1)0,5=2,04566906 𝐿=
ℎ × (𝑓0 + 2 × (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 2
𝐿=
0,3 × (1,632993162 + 2 × (1,838402084 + 2,04566906 + ⋯ + 3,515662235) 2 + 3,726779962) = 8,025683363
Metode simpson x 1 1,3 1,6 1,9 2,2 2,5 2,8 3,1 3,4 3,7 4 L
f(x)=(x+1)^1,5/(2x+1)^0,5 1,632993162 1,838402084 2,04566906 2,254116604 2,463361149 2,673169155 2,88339026 3,093923256 3,304697511 3,515662235 3,726779962 8,025510242
h=
𝑏−𝑎 10
=
4−1 10
= 0,3
f(1)=(1+1)1,5/(2*1+1)0,5=1,632993162 f(1,3)=(1,3+1)1,5/(2*1,3+1)0,5=1,838402084 f(1,6)=(1,6+1)1,5/(2*1,6+1)0,5=2,04566906 =
ℎ × (𝑓0 + 4 × (𝑓1 + 𝑓3 + ⋯ + 𝑓𝑔𝑎𝑛𝑗𝑖𝑙 ) + 2 × (𝑓2 + 𝑓4 + ⋯ + 𝑓𝑔𝑒𝑛𝑎𝑝 ) + 𝑓𝑛 ) 3
𝐿=
0,3 × (1,632993162 + 4 × (2,568965517 + ⋯ + 3,515662235) + 2 × (2,04566906 + ⋯ 3 + 3,304697511) + 3,726779962) = 8,025510242
5. F(X)=(2X+1)*3^X+2 Metode trapezoid dalam metode Trapezoid dan Simpson 1/3 ; n =10 Metode trapezoid X
F(X)=(2X+1)*3^X+2 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
11 14,70645558 19,69103954 26,35809377 35,23350066 47 62,54252367 83,00633896 109,8715581 145,0485463 191 128,8916113
L 𝑏−𝑎
h= 10 =
3−1 10
= 0,2
f(1)=(2*1+1)*31+2=11 f(1,2)=(2*1,2+1)*31,2+2=14,70645558 f(1,4)=(2*1,4+1)*31,4+2=19,69103954 𝐿=
ℎ × (𝑓0 + 2 × (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 2
𝐿=
0,2 × (11 + 2 × (14,70645558 + 19,69103954 + ⋯ + 145,0485463) + 191) 2 = 128,8916113
Metode snimpson X
F(X)=(2X+1)*3^X+2 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
11 14,70645558 19,69103954 26,35809377 35,23350066 47 62,54252367 83,00633896 109,8715581 145,0485463 191 128,0769988
L 𝑏−𝑎
h= 10 =
3−1 10
= 0,2
f(1)=(2*1+1)*31+2=11 f(1,2)=(2*1,2+1)*31,2+2=14,70645558 f(1,4)=(2*1,4+1)*31,4+2=19,69103954 =
ℎ × (𝑓0 + 4 × (𝑓1 + 𝑓3 + ⋯ + 𝑓𝑔𝑎𝑛𝑗𝑖𝑙 ) + 2 × (𝑓2 + 𝑓4 + ⋯ + 𝑓𝑔𝑒𝑛𝑎𝑝 ) + 𝑓𝑛 ) 3
𝐿=
0,2 × (11 + 4 × (14,70645558 + ⋯ + 145,0485463) + 2 × (19,69103954 + ⋯ 3 + 109,8715581) + 191) = 128,0769988
Selesaikan integral berikut dengan metode trafezoid dan metode simpson dengan n : 4 5
1. ∫2 𝑥 2 (√𝑥 − 2 ) 𝑏−𝑎 𝑛
h=
= 5-2/4 = 0,75
f(2) = 22 ( √2 − 2) = 0 f(2,75) = (2,75^3)*(2,75-2)^0,5 = 3,949386819 f(3,5) = (3,5^3)*(3,5-2)^0,5 =8,019507466 f(4,25) = (4,25^3)*(4,25-2)^0,5 =13,14239918 f(5) = (5^3)*(5-2)^0,5 = 19,36491673 X 2 2,75 3,5 4,25 5 -
f(x) = (x^3)*(x_2)^0,5 0 3,949386819 8,019507466 13,14239918 19,36491673
METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 0,3
L=(( 2 )) x(0+2*(3,949386819+8,019507466+13,14239918)+ 19,36491673) = 26,09531 -
METODE SIMPSON
-
L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn)
L = (0,3/3)*( 0+4*(3,949386819+13,14239918)+2*(8,019507466)+ 19,36491673) = 25,94277
2. f(x) = 3
∫ x^3 ∗ ln(2x + 1) 1
PENYELESAIAN: h = (b-a)/n = (3-1)/4 = 0,5 f(1) = (1^3)*(log(2*1+1)) = 0,477121255 f(1,5) = (1,5^3)*(log(2*1,5+1)) =2,031952471 f(2) = (2^3)*(log(2*2+1)) =5,591760035 f(2,5) = (2,5^3)*(log(2*2,5+1)) =12,15861329 f(3) = (3^3)*(log(2*3+1)) =22,81764708 X
f(x) = x^3*log(2x+1) 0,477121255 2,031952471 5,591760035 12,15861329 22,81764708
1 1,5 2 2,5 3 - METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) L=(0,2/2)(x(0,477121255+2*(2,031952471+5,591760035+12,15861329))+ 22,81764708) = 15,71485 METODE SIMPSON L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn) L = (0,2/3)*( 2,031952471+12,15861329)+2*(5,591760035+)+22,81764708) = 15,20676
3. f(x) = 5
∫(3^2x + 1)/(3x + 2) 3
PENYELESAIAN : h = (b-a)/10 = (5-3)/4 = 0,5 f(3) = (3^2*(3)+1)/(3*(3)+2) = 2,545454545 f(3,5) = (3^2*(3,5)+1)/(3*(3,5)+2) = 2,6 f(4) = (3^2*(4)+1)/(3*(4)+2) =2,642857143 f(4,5) = (3^2*(4,5)+1)/(3*(4,5)+2) =2,677419355 f(5) = (3^2*(5)+1)/(3*(5)+2) = 2,705882353 X 3 3,5 4 4,5 5 -
f(x) = (3^2x+1)/(3x+2) 2,545454545 2,6 2,642857143 2,677419355 2,705882353 METODE TRAFEZOID : ℎ
L = ((2 )) *(𝑓0 + 2 * (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) L=(0,2/2)(x(2,545454545+2*(2,6+2,642857143+2,677419355+)+2,705882353) = 5,272972 -
METODE SIMPSON L = (h/3)*(f0+4*( f1+f3+f4+f5+...fn-1.)+2*(f 2+f4+F6+f8....)+fn)
L = (0,2/3)*( 2,545454545+4*(2,6+2,677419355)+2*(2,642857143)+( 2,705882353) = 6,155407
Selesaikan integral berikut dengan metode trafezoid dan metode simpson dengan n : 10 4. f(x)=(x+1)^1,5/(2x+1)^0,5 Metode trapezoid dalam metode Trapezoid dan Simpson 1/3 ; n =10 metode trapezoid x 1 1,3 1,6 1,9 2,2 2,5 2,8 3,1 3,4 3,7 4 L 𝑏−𝑎
h=
10
=
f(x)=(x+1)^1,5/(2x+1)^0,5 1,632993162 1,838402084 2,04566906 2,254116604 2,463361149 2,673169155 2,88339026 3,093923256 3,304697511 3,515662235 3,726779962 8,025683363 4−1 10
=0,3
f(1)=(1+1)1,5/(2*1+1)0,5=1,632993162 f(1,3)=(1,3+1)1,5/(2*1,3+1)0,5=1,838402084 f(1,6)=(1,6+1)1,5/(2*1,6+1)0,5=2,04566906 𝐿=
ℎ × (𝑓0 + 2 × (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 2
𝐿=
0,3 × (1,632993162 + 2 × (1,838402084 + 2,04566906 + ⋯ + 3,515662235) 2 + 3,726779962) = 8,025683363
Metode simpson x 1 1,3 1,6 1,9 2,2 2,5 2,8 3,1 3,4 3,7 4 L
f(x)=(x+1)^1,5/(2x+1)^0,5 1,632993162 1,838402084 2,04566906 2,254116604 2,463361149 2,673169155 2,88339026 3,093923256 3,304697511 3,515662235 3,726779962 8,025510242
h=
𝑏−𝑎 10
=
4−1 10
= 0,3
f(1)=(1+1)1,5/(2*1+1)0,5=1,632993162 f(1,3)=(1,3+1)1,5/(2*1,3+1)0,5=1,838402084 f(1,6)=(1,6+1)1,5/(2*1,6+1)0,5=2,04566906 =
ℎ × (𝑓0 + 4 × (𝑓1 + 𝑓3 + ⋯ + 𝑓𝑔𝑎𝑛𝑗𝑖𝑙 ) + 2 × (𝑓2 + 𝑓4 + ⋯ + 𝑓𝑔𝑒𝑛𝑎𝑝 ) + 𝑓𝑛 ) 3
𝐿=
0,3 × (1,632993162 + 4 × (2,568965517 + ⋯ + 3,515662235) + 2 × (2,04566906 + ⋯ 3 + 3,304697511) + 3,726779962) = 8,025510242
5. F(X)=(2X+1)*3^X+2 Metode trapezoid dalam metode Trapezoid dan Simpson 1/3 ; n =10 Metode trapezoid X
F(X)=(2X+1)*3^X+2 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
11 14,70645558 19,69103954 26,35809377 35,23350066 47 62,54252367 83,00633896 109,8715581 145,0485463 191 128,8916113
L 𝑏−𝑎
h= 10 =
3−1 10
= 0,2
f(1)=(2*1+1)*31+2=11 f(1,2)=(2*1,2+1)*31,2+2=14,70645558 f(1,4)=(2*1,4+1)*31,4+2=19,69103954 𝐿=
ℎ × (𝑓0 + 2 × (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛−1 ) + 𝑓𝑛 ) 2
𝐿=
0,2 × (11 + 2 × (14,70645558 + 19,69103954 + ⋯ + 145,0485463) + 191) 2 = 128,8916113
Metode snimpson X
F(X)=(2X+1)*3^X+2 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6 2,8 3
11 14,70645558 19,69103954 26,35809377 35,23350066 47 62,54252367 83,00633896 109,8715581 145,0485463 191 128,0769988
L 𝑏−𝑎
h= 10 =
3−1 10
= 0,2
f(1)=(2*1+1)*31+2=11 f(1,2)=(2*1,2+1)*31,2+2=14,70645558 f(1,4)=(2*1,4+1)*31,4+2=19,69103954 =
ℎ × (𝑓0 + 4 × (𝑓1 + 𝑓3 + ⋯ + 𝑓𝑔𝑎𝑛𝑗𝑖𝑙 ) + 2 × (𝑓2 + 𝑓4 + ⋯ + 𝑓𝑔𝑒𝑛𝑎𝑝 ) + 𝑓𝑛 ) 3
𝐿=
0,2 × (11 + 4 × (14,70645558 + ⋯ + 145,0485463) + 2 × (19,69103954 + ⋯ 3 + 109,8715581) + 191) = 128,0769988
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