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GATE CH 2020 (with Solutions).pdf

Q67

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Consider the following data se t./x1351525 f ( x)6 8 12 5 10/rd
²5()²⁵()
Calculate the value of f ( x )dx by Simpson 1/3 method _______.
11

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Answer

242 ()ʳ^d

Solution

Simpson 1/3 method
f(x)ₐ^ᵇf(x)
dx = h{/3(y_0+y_n)+4(y_1+y_3+−−−−−+y_n)+2(y_2+y_4+−−−−+y_n_{−}_2)} /text{Where}
h=((ba)/frac/frac)nh/textrepresents/textequal/textintervalbwb/textanda125f(x))h = ((b-a) {/frac{/frac{)}{nh} /text{represents} /text{equal} /text{interval} b}{w} b /text{and} a}_1^{25} f (x))
dx=15f(x)dx+525f(x)dxdx = ₁^5f(x)dx+₅^{25}f(x)dx
h=/frac(51)2=2⇒h = /frac{(5-1)}{2} = 2
h=/frac(255)2=10⇒h = /frac{(25-5)}{2} = 10
5f(x)₁^5f(x)
dx=h/3(y0+yn)+4(y1)+2(0)=2/36+10+4/times8+2/times0=/frac23/times48=32dx = h{/3(y_0+y_n)+4(y_1)+2(0)} = 2{/36+10+4 /times 8+2 /times 0} = /frac{2}{3} /times 48 = 32
25f(x)₅^{25}f(x)
dx=h/3(y0+yn)+4(y1)+2(0)=10/310+5+4/times12+2/times0=/frac103/times63=210dx = h{/3(y_0+y_n)+4(y_1)+2(0)} = 10{/310+5+4 /times 12+2 /times 0} = /frac{10}{3} /times 63 = 210
∴ Adding this two integral :
25f(x)dx=32+210=242125f(x)dx = 32+210 = 2421
ϖϖϖϖ

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