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Evaluate $\int_{-1}^{1}\frac{1}{1+x^2}dx$ by using (i)Trapezoidal rule (ii) Simpon's 1/3 rule (iii) Simpson's 3/8 rule .
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$ h = \frac { 1-(-1)}{6} = \frac{2}{6} = \frac{1}{3} \\ $

X -1 -2/3 -1/3 0 1/3 2/3 1
Y 0.5 0.6923 0.9 1 0.9 0.6923 0.5
y0 y1 y2 y3 y4 y5 y6

$ \text{1) Trapezoidal Rule } \\ $

$ I = \frac{h}{2} ( X + 2R ) = \frac{h}{2} \left[(y_0+y_6) + 2(y_1+y_2+y_3+y_4+y_5) \right ] \\ $

$ = \frac{1}{3*2} \left[(0.5 + 0.5) + 2( 0.6923 + 0.9 + 1 + 0.9 + 0.6923 ) \right ] \\ $

$ = \frac{1}{6} \left[ 9.3692 \right ] = 1.5615 \\ $

$ \text{2) Simpson's 1/3 rd Rule } \\ $

$ I = \frac{h}{3} [ X + 2E + 40 ] \\ $

$ = \frac{h}{3} [ (y_0+ y_6) + 2(y_2+y_4 ) + 4 (y_1 + y_3 + y_5 ) ] \\ $

$ = \frac{1}{3*3} [ (0.5 + 0.5) + 2(0.9 + 0.9) + 4(0.6923 + 1 + 0.6923 ) ] \\ $

$ = \frac{1}{9} [ 14. 1384] = 1.5709 \\ $

$ \text{ 3) Simpson's 3/8 th Rule } \\ $

$ I = \frac{3h}{8}[ X +2T + 3R ] \\ $

$ = 3* \frac{1}{3*8} [(y_0 + y_6) + 2(y_3) + 3(y_1 + y_2 + y_4 + y_5) ] \\ $

$ = \frac{1}{8} [ 12.5538 ] = 1.5692 \\ $

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