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 ) / 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: The given integral is ∫ 2 10 ( 4 x + 3 ) d x . So, the function is f ( x ) = 4 x + 3 . The given interval is [ a , b ] = [ 2 , 10 ] and 2 n = 4 . Divide the interval [ 2 , 10 ] into four equal subintervals. So, the length of each subinterval becomes, Δ x = ( b − a ) / 2 n = ( 10 − 2 ) / 4 = 8 / 4 = 2 Thus, the length of each interval is 2. So, the end points of the each subinterval are x 0 = 2 , x 1 = 4 , x 2 = 6 , x 3 = 8 and x 4 = 10 . Calculate the value of the function f ( x ) = ( 4 x + 3 ) at each end point as follows. x f ( x ) 2 ( 4 x + 3 ) = ( 4 ( 2 ) + 3 ) = 11 4 ( 4 x + 3 ) = ( 4 ( 4 ) + 3 ) = 19 6 ( 4 x + 3 ) = ( 4 ( 6 ) + 3 ) = 27 8 ( 4 x + 3 ) = ( 4 ( 8 ) + 3 ) = 35 10 ( 4 x + 3 ) = ( 4 ( 10 ) + 3 ) = 43 Table 1 Calculate S 4 by using Simpson’s rule

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Numerical Integration

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Chapter 6.4, Problem 35E

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To show: The approximate value of the integral ∫210(4x+3)dx using the Simpson’s rule is same as the exact value of ∫210(4x+3)dx using fundamental theorem of calculus.

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Chapter 6.4, Problem 34E

Chapter 6.4, Problem 36E

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The approximate value of the integral ∫ 2 10 ( 4 x + 3 ) d x using the Simpson’s rule is same as the exact value of ∫ 2 10 ( 4 x + 3 ) d x using fundamental theorem of calculus.

Solution Summary: The author compares the approximate value of the integral with the exact value using the Simpson's rule. The function is f(x)=4x+3.

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To show: The approximate value of the integral ∫210(4x+3)dx using the Simpson’s rule is same as the exact value of ∫210(4x+3)dx using fundamental theorem of calculus.

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