Explanation Formula used: Simpson’s rule: Let f be a function on an interval [ a , b ] . Partition [ a , b ] into 2 n subintervals of equal length Δ x = ( b − a ) / 2 n with end points a = x 0 < x 1 < x 2 < ... < x n = b . Then S 2 n = [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) .... + 4 f ( x 2 n − 1 ) + f ( x 2 n ) ] Δ x 3 is an approximation of ∫ a b f ( x ) d x . Calculation: Here, the given function f ( x ) = x 2 + 3 x + 5 , [ a , b ] = [ − 1 , 1 ] and 2 n = 2 . Thus, partition the interval [ − 1 , 1 ] into 2 equal subintervals and the length of the each interval is Δ x = ( b − a ) 2 n . Δ x = ( b − a ) 2 n = 1 − ( − 1 ) 2 = 2 2 = 1 Thus, the length of each interval is 1 and the end points of the each subinterval are x 0 = − 1 , x 1 = 0 and x 2 = 1 . The corresponding values of the function f ( x ) = x 2 + 3 x + 5 at each end point are as shown below in Table 1. x f ( x ) − 1 x 2 + 3 x + 5 = ( − 1 ) 2 + 3 ( − 1 ) + 5 = 3 0 x 2 + 3 x + 5 = ( 0 ) 2 + 3 ( 0 ) + 5 = 5 1 x 2 + 3 x + 5 = ( 1 ) 2 + 3 ( 1 ) + 5 = 9 Table 1 By Simpson’s rule, obtain the value of the integral ∫ − 1 1 ( x 2 + 3 x + 5 ) d x as follows

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Chapter 6.4, Problem 37E

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To show: The Simpson’s rule with n=1 gives the exact value of the integral ∫−11(x2+3x+5)dx.

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Chapter 6.4, Problem 36E

Chapter 6.4, Problem 38E

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The Simpson’s rule with n = 1 gives the exact value of the integral ∫ − 1 1 ( x 2 + 3 x + 5 ) d x .

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To show: The Simpson’s rule with n=1 gives the exact value of the integral ∫−11(x2+3x+5)dx.

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