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By the application of Simpson's one-third rule numerical integration, subintervals, the value of int_(0)^(1) (dx)/(1x)is |
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Answer» `(17)/(36)` ie, `h=(1-0)/(2)=(1)/(2)` `therefore underset(0)overset(1)INT (1)/(1+x)DX=(h)/(3)[(y_(0)+y_(2))+4(y_(1))]` `"At" x=0, y=1` `x=(1)/(2), y_(1)=(2)/(3)` `and x=1, y_(2)=(1)/(2)` `therefore underset(0)overset(1)int (1)/(1+x)dx=(1)/(2.3)[(1+(1)/2)+4((2)/(3))]` `=(1)/(6)[(3)/(2)+(8)/(3)]=(25)/(36)` |
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