Explanation Given information: The integral function is ∫ 0 2 x 4 d x . The upper limit is 2 and lower limit is 0. The number of sub intervals as n = 6 . Calculation: Trapezoidal rule: Calculate the approximate value of integral by using Trapezoidal rule. Show the integral function as follows: ∫ 0 2 x 4 d x (1) Show the Trapezoidal rule as follows: ∫ a b f ( x ) d x ≈ T n = Δ x 2 [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ... + 2 f ( x n − 1 ) + f ( x n ) ] Here, T n is trapezoidal approximation, x 1 , x 2 , ... x n are subintervals, Δ x = b − a n and x i = a + i Δ x . Calculate the width of the subinterval ( Δ x ) using the formula: Δ x = b − a n Here, b is upper limit, a is lower limit, and n is number of subintervals. Substitute 2 for b , 0 for a , and 6 for n . Δ x = 2 − 0 6 = 1 3 Hence, the subinterval is ( 1 3 ) for the limits [ 0 , 2 ] . The subinterval values are 0, ( 1 3 ) , ( 2 3 ) , 1, ( 4 3 ) , ( 5 3 ) , and 2 for x 0 , x 1 , x 2 , x 3 , x 4 , x 5 , and x 6 . Apply Trapezoidal rule in Equation (1). ∫ 0 2 x 4 d x ≈ T n = Δ x 2 [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ... + 2 f ( x n − 1 ) + f ( x n ) ] Substitute ( 1 3 ) for Δ x and 6 for n . T 6 = 1 6 [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ... + 2 f ( x 5 ) + f ( x 6 ) ] (2) Apply the subinterval values in Equation (2). T 6 = 1 6 [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ... + 2 f ( x 5 ) + f ( x 6 ) ] = 1 6 [ f ( 0 ) + 2 f ( 1 3 ) + 2 f ( 2 3 ) + 2 f ( 1 ) + 2 f ( 4 3 ) + 2 f ( 5 3 ) + f ( 2 ) ] (3) Apply f ( x ) = x 4 in Equation (3). T 6 = 1 6 [ f ( 0 ) + 2 f ( 1 3 ) + 2 f ( 2 3 ) + 2 f ( 1 ) + 2 f ( 4 3 ) + 2 f ( 5 3 ) + f ( 2 ) ] = 1 6 [ 0 + 2 ( 1 3 ) 4 + 2 ( 2 3 ) 4 + 2 ( 1 ) 4 + 2 ( 4 3 ) 4 + 2 ( 5 3 ) 4 + ( 2 ) 4 ] ≈ 6.695473 Therefore, the approximate value of integral by using Trapezoidal rule is 6.695473. Midpoint rule: Calculate the approximate value of integral by using Midpoint rule. Show the Midpoint rule as follows: ∫ a b f ( x ) d x ≈ M n = Δ x [ f ( x ¯ 1 ) + f ( x ¯ 2 ) + ... + f ( x ¯ n ) ] Consider the condition for midpoint rule as follows: Δ x = b − a n x ¯ i = 1 2 ( x i − 1 + x i ) = Midpoint of [ x i − 1 , x i ] Here, M n is midpoint approximation, x ¯ 1 , x ¯ 2 , ... x ¯ n are average subintervals, b is upper limit, a is lower limit, and n is number of subintervals. Apply midpoint rule in Equation (1). ∫ 0 2 x 4 d x ≈ M n = Δ x [ f ( x ¯ 1 ) + f ( x ¯ 2 ) + ... + f ( x ¯ n ) ] Substitute ( 1 3 ) for Δ x and 6 for n . M 6 = ( 1 3 ) [ f ( x ¯ 1 ) + f ( x ¯ 2 ) + ... + f ( x ¯ 6 ) ] (4) Find the average subinterval using the formula. x ¯ i = 1 2 ( x i − 1 + x i ) x ¯ 1 = 1 2 ( x 0 + x 1 ) = 1 2 ( 0 + 1 3 ) = 1 6 Similarly calculate the values for x ¯ 2 , x ¯ 3 ... x ¯ n . Modify Equation (4). M 6 = ( 1 3 ) [ f ( x ¯ 1 ) + f ( x ¯ 2 ) + ... + f ( x ¯ 6 ) ] = ( 1 3 ) [ f ( x 0 + x 1 2 ) + f ( x 1 + x 2 2 ) + + ... + f ( x 5 + x 6 2 ) ] (5) Apply the subinterval values in Equation (5). M 6 = ( 1 3 ) [ f ( x 0 + x 1 2 ) + f ( x 1 + x 2 2 ) + + ... + f ( x 5 + x 6 2 ) ] = ( 1 3 ) [ f ( 1 6 ) + f ( 3 6 ) + f ( 5 6 ) + f ( 7 6 ) + f ( 9 6 ) + f ( 11 6 ) ] (6) Apply f ( x ) = x 4 in Equation (6). M 6 = ( 1 3 ) [ ( 1 6 ) 4 + ( 3 6 ) 4 + ( 5 6 ) 4 + ( 7 6 ) 4 + ( 9 6 ) 4 + ( 11 6 ) 4 ] = 6.252572 Therefore, the approximate value of integral by using Midpoint rule is 6.252572. Simpson’s rule: Calculate the approximate value of integral by using Simpson’s rule. Show the Simpson’s rule as follows: ∫ a b f ( x ) d x ≈ S n = Δ x 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) ... + 2 f ( x n − 2 ) + 4 f ( x n − 1 ) + f ( x n ) ] Consider the condition for Simpson rule as follows: Δ x = b − a n Number of subinterval n is even. Here, S n is Simpson’s approximation, x 1 , x 2 , ... x n are subintervals, b is upper limit, a is lower limit, and n is number of subintervals. Apply Simpson’s rule in Equation (1). ∫ 0 2 x 4 d x ≈ S n = Δ x 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) ... + 2 f ( x n − 2 ) + 4 f ( x n − 1 ) + f ( x n ) ] Substitute ( 1 3 ) for Δ x and 6 for n . S 6 = 1 3 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) ... + 2 f ( x 8 ) + 4 f ( x 9 ) + f ( x 10 ) ] (7) Apply the subinterval values in Equation (7). S 6 = 1 9 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) ... + 2 f ( x 8 ) + 4 f ( x 9 ) + f ( x 10 ) ] = 1 9 [ f ( 0 ) + 4 f ( 1 3 ) + 2 f ( 2 3 ) + 4 f ( 3 3 ) + 2 f ( 4 3 ) + 4 f ( 5 3 ) + f ( 2 ) ] ...... (8) Apply f ( x ) = x 4 in Equation (8). S 6 = 1 9 [ f ( 0 ) + 4 f ( 1 3 ) + 2 f ( 2 3 ) + 4 f ( 3 3 ) + 2 f ( 4 3 ) + 4 f ( 5 3 ) + f ( 2 ) ] = 1 9 [ ( 0 4 ) + 4 ( 1 3 ) 4 + 2 ( 2 3 ) 4 + 4 ( 3 3 ) 4 + 2 ( 4 3 ) 4 + 4 ( 5 3 ) 4 + ( 2 ) 4 ] = 6.403292 Therefore, the approximate value of integral by using Simpson’s rule is6.403292. Find the actual value of the integral as follows: Evaluate the integral in Equation (1). ∫ 0 2 x 4 d x = [ x 5 5 ] 0 2 = [ 2 5 5 − 0 ] = 32 5 − 0 = 6.4 Compare the integral value obtained from Trapezoidal rule and actual value. Find the actual error in trapezoidal rule ( E T ) as follows: E T = Actual value − T 6 Substitute 6.4 for actual value and 6.695473for T 6 . E T = Actual value − T 6 = 6.4 − 6.695473 = − 0.295473 Therefore, the actual error in Trapezoidal ruleis − 0.295473 _ . Compare the integral value obtained from Midpoint rule and actual value. Find the actual error value in Midpoint rule ( E M ) . E M = Actual value − M 6 Substitute 6.4 for actual value and 6.252572 for M 6 . E M = Actual value − M 6 = 6.4 − 6.252572 = 0.147428 Therefore, the actual error in Midpoint rule is 0.147428 _ . Compare the integral value obtained from Simpson’s rule and actual value. Find the actual error value in Simpson’s rule ( E S ) . E S = Actual value − S 6 Substitute 6.4 for actual value and 6.403292 for S 6 . E S = Actual value − S 6 = 6.4 − 6.403292 = − 0.003292 Therefore, the actual error in Simpson’s rule is − 0.003292 _ . For number of subinterval n = 12 , Trapezoidal rule: Calculate the approximate value of integral by using Trapezoidal rule. Show the integral function as follows: ∫ 0 2 x 4 d x (1) Show the Trapezoidal rule as follows: ∫ a b f ( x ) d x ≈ T n = Δ x 2 [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + ... + 2 f ( x n − 1 ) + f ( x n ) ] Here, T n is trapezoidal approximation, x 1 , x 2 ,

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Chapter 7.7, Problem 27E

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To Calculate: the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n=6 and n=12.

To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.

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Chapter 7.7, Problem 26E

Chapter 7.7, Problem 28E

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the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n = 6 and n = 12 . To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.

Solution Summary: The author calculates the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson's rule.

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To Calculate: the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n=6 and n=12.

To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.

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Chapter 7 Solutions

the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n = 6 and n = 12 . To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.

Solution Summary: The author calculates the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson's rule.

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To Calculate: the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n=6 and n=12. To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.

Want to see the full answer?

Chapter 7 Solutions

To Calculate: the approximate value of the integral using Trapezoidal rule, Midpoint rule and Simpson’s rule for the number of subintervals n=6 and n=12.

To Find: The actual error value in Trapezoidal rule, Midpoint rule and Simpson’s rule.