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The integral of ∫ 0 1 sin 1 2 π x d x by Trapezoidal rule for different values of h . The exact value is J = 0.63662 _ . The approximation value for h = 1 is ∫ 0 1 sin 1 2 π x d x ≈ 0.5 _ . The error value of the integral for h = 1 is 0.03307 _ . The approximation value for h = 0.5 is ∫ 0 1 sin 1 2 π x d x ≈ 0.603553 _ . The error value of the integral for h = 0.5 is 0.03307 _ . The approximation value for h = 0.25 is ∫ 0 1 sin 1 2 π x d x ≈ 0.6282765 _ . The error value of the integral for h = 0.25 is 0.0083 _ . The estimated error for h = 1 is ε h 2 ≈ 0.034518 _ . The estimated error for h = 0.5 is ε h 2 ≈ 0.0082 _ . The given integral equation is: J = ∫ 0 1 sin 1 2 π x d x Now, solve the integral J = ∫ 0 1 x 4 d x and obtain the exact solution: J = ∫ 0 1 sin 1 2 π x d x = [ − 2 π cos 1 2 π x ] 0 1 = − 2 π [ cos 1 2 π − cos 0 ] = 0.63662 Hence, the exact value is J = 0.63662 _ . Take h = 1 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 1 y = f ( x ) 0 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 ] ≈ 1 2 [ 1 ] ≈ 0.5 Therefore, the approximation value for h = 1 is ∫ 0 1 sin 1 2 π x d x ≈ 0.5 _ . Now, calculate the error: Error = True value − Approximation value = 0.63662 − 0.5 = 0.03307 Hence, the error value of the integral for h = 1 is 0.03307 _ . Take h = 0.5 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 0.5 1 y = f ( x ) 0 0.707106 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 + 2 ( 0.707106 ) ] ≈ 0.5 [ 1.207106 ] ≈ 0.603553 Therefore, the approximation value for h = 0.5 is ∫ 0 1 sin 1 2 π x d x ≈ 0.603553 _ . Now, calculate the error: Error = True value − Approximation value = 0.63662 − 0.60355 = 0.03307 Hence, the error value of the integral for h = 0.5 is 0.03307 _ . Take h = 0.25 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 0.25 0.5 0.75 1 y = f ( x ) 0 0.3824 0.707106 0.9236 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 + 2 ( 0.3824 + 0.707106 + 0.9236 ) ] ≈ 0.25 [ 2.513106 ] ≈ 0.6282765 Therefore, the approximation value for h = 0.25 is ∫ 0 1 sin 1 2 π x d x ≈ 0.6282765 _ . Now, calculate the error: Error = True value − Approximation value = 0.62662 − 0.6282765 = 0.0083 Hence, the error value of the integral for h = 0.25 is 0.0083 _ . Estimate the error by using the formula. ε h 2 ≈ 1 3 [ J h 2 − J h ] Substitute the value h = 1 and J = 0.5 in ε h 2 ≈ 1 3 [ J h 2 − J h ] . ε h 2 ≈ 1 3 [ J h 2 − J h ] ≈ 1 3 [ 0.603553 − 0.5 ] ≈ 0.034518 Therefore, the estimated error for h = 1 is ε h 2 ≈ 0.034518 _ . Substitute the value h = 0.5 and J = 0.28125 in ε h 2 ≈ 1 3 [ J h 2 − J h ] . ε h 2 ≈ 1 3 [ J h 2 − J h ] ≈ 1 3 [ 0.6282765 − 0.603553 ] ≈ 0.0082 Therefore, the estimated error for h = 0.5 is ε h 2 ≈ 0.0082 _ .

The integral of ∫ 0 1 sin 1 2 π x d x by Trapezoidal rule for different values of h . The exact value is J = 0.63662 _ . The approximation value for h = 1 is ∫ 0 1 sin 1 2 π x d x ≈ 0.5 _ . The error value of the integral for h = 1 is 0.03307 _ . The approximation value for h = 0.5 is ∫ 0 1 sin 1 2 π x d x ≈ 0.603553 _ . The error value of the integral for h = 0.5 is 0.03307 _ . The approximation value for h = 0.25 is ∫ 0 1 sin 1 2 π x d x ≈ 0.6282765 _ . The error value of the integral for h = 0.25 is 0.0083 _ . The estimated error for h = 1 is ε h 2 ≈ 0.034518 _ . The estimated error for h = 0.5 is ε h 2 ≈ 0.0082 _ . The given integral equation is: J = ∫ 0 1 sin 1 2 π x d x Now, solve the integral J = ∫ 0 1 x 4 d x and obtain the exact solution: J = ∫ 0 1 sin 1 2 π x d x = [ − 2 π cos 1 2 π x ] 0 1 = − 2 π [ cos 1 2 π − cos 0 ] = 0.63662 Hence, the exact value is J = 0.63662 _ . Take h = 1 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 1 y = f ( x ) 0 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 ] ≈ 1 2 [ 1 ] ≈ 0.5 Therefore, the approximation value for h = 1 is ∫ 0 1 sin 1 2 π x d x ≈ 0.5 _ . Now, calculate the error: Error = True value − Approximation value = 0.63662 − 0.5 = 0.03307 Hence, the error value of the integral for h = 1 is 0.03307 _ . Take h = 0.5 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 0.5 1 y = f ( x ) 0 0.707106 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 + 2 ( 0.707106 ) ] ≈ 0.5 [ 1.207106 ] ≈ 0.603553 Therefore, the approximation value for h = 0.5 is ∫ 0 1 sin 1 2 π x d x ≈ 0.603553 _ . Now, calculate the error: Error = True value − Approximation value = 0.63662 − 0.60355 = 0.03307 Hence, the error value of the integral for h = 0.5 is 0.03307 _ . Take h = 0.25 . Apply the Trapezoidal rule: ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] Form the function, find the values of integral and shows in the table given below: x 0 0.25 0.5 0.75 1 y = f ( x ) 0 0.3824 0.707106 0.9236 1 Substitute the values from the table in ∫ a b f ( x ) d x ≈ h 2 [ f ( a ) + f ( b ) + 2 ( f ( x 1 ) + f ( x 2 ) + ⋅ ⋅ ⋅ + f ( x n − 1 ) ) ] . ∫ 0 1 sin 1 2 π x d x ≈ h 2 [ 0 + 1 + 2 ( 0.3824 + 0.707106 + 0.9236 ) ] ≈ 0.25 [ 2.513106 ] ≈ 0.6282765 Therefore, the approximation value for h = 0.25 is ∫ 0 1 sin 1 2 π x d x ≈ 0.6282765 _ . Now, calculate the error: Error = True value − Approximation value = 0.62662 − 0.6282765 = 0.0083 Hence, the error value of the integral for h = 0.25 is 0.0083 _ . Estimate the error by using the formula. ε h 2 ≈ 1 3 [ J h 2 − J h ] Substitute the value h = 1 and J = 0.5 in ε h 2 ≈ 1 3 [ J h 2 − J h ] . ε h 2 ≈ 1 3 [ J h 2 − J h ] ≈ 1 3 [ 0.603553 − 0.5 ] ≈ 0.034518 Therefore, the estimated error for h = 1 is ε h 2 ≈ 0.034518 _ . Substitute the value h = 0.5 and J = 0.28125 in ε h 2 ≈ 1 3 [ J h 2 − J h ] . ε h 2 ≈ 1 3 [ J h 2 − J h ] ≈ 1 3 [ 0.6282765 − 0.603553 ] ≈ 0.0082 Therefore, the estimated error for h = 0.5 is ε h 2 ≈ 0.0082 _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Advanced Engineering Mathematics

Ch. 19.1 - Floating point. Write 84.175, −528.685,...Ch. 19.1 - Write −76.437125, 60100, and −0.00001 in...Ch. 19.1 - Prob. 3P Ch. 19.1 - Order of terms, in adding with a fixed number of...Ch. 19.1 - Prob. 5P Ch. 19.1 - Nested form. Evaluate at x = 3.94 using 3S...Ch. 19.1 - Quadratic equation. Solve x2 − 30x + 1 = 0 by (4)...Ch. 19.1 - Solve x2 − 40x + 2 = 0, using 4S-computation. Ch. 19.1 - Prob. 9P Ch. 19.1 - Instability. For small |a| the equation (x − k)2 =...

Ch. 19.1 - (a) In addition and subtraction, a bound for the...Ch. 19.1 - Prob. 12P Ch. 19.1 - (b) In multiplication and division, an error bound...Ch. 19.1 - Prob. 14P Ch. 19.1 - Prob. 15P Ch. 19.1 - Prob. 16P Ch. 19.1 - Prob. 17P Ch. 19.1 - Prob. 18P Ch. 19.1 - Prob. 19P Ch. 19.1 - Prob. 20P Ch. 19.1 - Prob. 21P Ch. 19.1 - Prob. 22P Ch. 19.1 - Prob. 23P Ch. 19.1 - Prob. 24P Ch. 19.1 - Prob. 27P Ch. 19.1 - Prob. 28P Ch. 19.1 - Prob. 29P Ch. 19.1 - Prob. 30P Ch. 19.2 - Prob. 1P Ch. 19.2 - Prob. 2P Ch. 19.2 - Prob. 3P Ch. 19.2 - Prob. 4P Ch. 19.2 - Prob. 5P Ch. 19.2 - Prob. 6P Ch. 19.2 - Prob. 7P Ch. 19.2 - Prob. 8P Ch. 19.2 - Prob. 9P Ch. 19.2 - Prob. 10P Ch. 19.2 - Prob. 11P Ch. 19.2 - Prob. 13P Ch. 19.2 - Prob. 14P Ch. 19.2 - Prob. 15P Ch. 19.2 - Prob. 16P Ch. 19.2 - Prob. 17P Ch. 19.2 - Prob. 18P Ch. 19.2 - Prob. 19P Ch. 19.2 - Prob. 20P Ch. 19.2 - Prob. 21P Ch. 19.2 - Prob. 22P Ch. 19.2 - Prob. 23P Ch. 19.2 - Prob. 26P Ch. 19.2 - Prob. 27P Ch. 19.2 - Prob. 28P Ch. 19.2 - Prob. 29P Ch. 19.3 - Prob. 1P Ch. 19.3 - Prob. 2P Ch. 19.3 - Prob. 3P Ch. 19.3 - Prob. 4P Ch. 19.3 - Prob. 5P Ch. 19.3 - Prob. 6P Ch. 19.3 - Prob. 7P Ch. 19.3 - Prob. 8P Ch. 19.3 - Prob. 9P Ch. 19.3 - Prob. 10P Ch. 19.3 - Prob. 11P Ch. 19.3 - Prob. 12P Ch. 19.3 - Prob. 13P Ch. 19.3 - Prob. 14P Ch. 19.3 - Prob. 15P Ch. 19.3 - Prob. 16P Ch. 19.3 - Prob. 17P Ch. 19.3 - Prob. 18P Ch. 19.4 - Prob. 2P Ch. 19.4 - Prob. 3P Ch. 19.4 - Prob. 4P Ch. 19.4 - Prob. 5P Ch. 19.4 - Prob. 6P Ch. 19.4 - Prob. 7P Ch. 19.4 - Prob. 8P Ch. 19.4 - Prob. 9P Ch. 19.4 - Prob. 10P Ch. 19.4 - Prob. 11P Ch. 19.4 - Prob. 12P Ch. 19.4 - Prob. 13P Ch. 19.4 - Prob. 14P Ch. 19.4 - Prob. 15P Ch. 19.4 - Prob. 16P Ch. 19.4 - Prob. 17P Ch. 19.4 - Prob. 19P Ch. 19.5 - Prob. 1P Ch. 19.5 - Prob. 2P Ch. 19.5 - Prob. 3P Ch. 19.5 - Prob. 4P Ch. 19.5 - Prob. 5P Ch. 19.5 - Prob. 6P Ch. 19.5 - Prob. 7P Ch. 19.5 - Prob. 8P Ch. 19.5 - Prob. 9P Ch. 19.5 - Prob. 10P Ch. 19.5 - Prob. 11P Ch. 19.5 - Prob. 12P Ch. 19.5 - Prob. 13P Ch. 19.5 - Prob. 14P Ch. 19.5 - Prob. 15P Ch. 19.5 - Prob. 16P Ch. 19.5 - Prob. 17P Ch. 19.5 - Prob. 18P Ch. 19.5 - Prob. 19P Ch. 19.5 - Prob. 20P Ch. 19.5 - Prob. 21P Ch. 19.5 - Prob. 22P Ch. 19.5 - Prob. 23P Ch. 19.5 - Prob. 24P Ch. 19.5 - Prob. 25P Ch. 19.5 - Prob. 27P Ch. 19.5 - Prob. 28P Ch. 19.5 - Prob. 29P Ch. 19.5 - Prob. 30P Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ Ch. 19 - Prob. 11RQ Ch. 19 - Prob. 12RQ Ch. 19 - Prob. 13RQ Ch. 19 - Prob. 14RQ Ch. 19 - Prob. 15RQ Ch. 19 - Prob. 16RQ Ch. 19 - Prob. 17RQ Ch. 19 - Prob. 18RQ Ch. 19 - Prob. 19RQ Ch. 19 - Prob. 20RQ Ch. 19 - Prob. 21RQ Ch. 19 - Prob. 22RQ Ch. 19 - Prob. 23RQ Ch. 19 - Prob. 24RQ Ch. 19 - Prob. 25RQ Ch. 19 - Prob. 26RQ Ch. 19 - Prob. 27RQ Ch. 19 - Prob. 28RQ Ch. 19 - Prob. 29RQ Ch. 19 - Prob. 30RQ Ch. 19 - Prob. 31RQ Ch. 19 - Prob. 32RQ Ch. 19 - Prob. 33RQ Ch. 19 - Prob. 34RQ Ch. 19 - Prob. 35RQ

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