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Math Advanced Math ADVANCED ENGINEERING MATHEMATICS

The mean and variance of the random variable X with probability density f ( x ) . The mean is μ = 1 4 _ and the variance is σ 2 = 1 16 _ . Formula used: Mean of X : μ = E ( X ) = ∫ − ∞ ∞ x f ( x ) Variance of X : σ 2 = E ( [ X − μ ] 2 ) = ∫ − ∞ ∞ ( x − μ ) 2 f ( x ) Integration by parts: Let u and v be two functions, then integration by parts states that ∫ u v d x = u v 1 − u ' v 2 + u ' ' v 3 − u ' ' ' v 4 + … . Calculation: It is given that the probability density function is f ( x ) = 4 e − 4 x ( x ≥ 0 ) . Calculate the mean as shown below. μ = ∫ 0 ∞ x 4 e − 4 x d x = 4 ∫ 0 ∞ x e − 4 x d x Consider u = x and v = e − 4 x and apply integration by parts to solve the above integration as follows. μ = 4 [ ( x e − 4 x − 4 ) 0 ∞ − ( e − 4 x 16 ) 0 ∞ ] = 4 [ 0 + 1 16 ] = 1 4 Thus, the value of mean is, μ = 1 4 . Now, calculate the variance as shown below. σ 2 = ∫ 0 ∞ ( x − 1 4 ) 2 4 e − 4 x d x = 4 ∫ 0 ∞ ( x − 1 4 ) 2 e − 4 x d x = 4 ∫ 0 ∞ ( x 2 − x 2 + 1 16 ) e − 4 x d x = 4 ∫ 0 ∞ ( x 2 e − 4 x − x e − 4 x 2 + e − 4 x 16 ) d x = 4 ( ∫ 0 ∞ x 2 e − 4 x ︸ term 1 − ∫ 0 ∞ x e − 4 x 2 ︸ term 2 + ∫ 0 ∞ e − 4 x 16 ︸ term 3 ) d x Simplify term 1 by using integration by parts, as follows. Consider u = x 2 and v = e − 4 x . ∫ 0 ∞ e − 4 x x 2 d x = ( x 2 e − 4 x − 4 − 2 x e − 4 x 16 + 2 e − 4 x − 64 ) 0 ∞ = ( x 2 e − 4 x − 4 − x e − 4 x 8 + e − 4 x − 32 ) 0 ∞ = [ 8 x 2 e − 4 x + 4 x e − 4 x + e − 4 x − 32 ] 0 ∞ = [ 1 − 32 e − 4 x ( 8 x 2 + 4 x + 1 ) ] 0 ∞ = − 1 32 Simplify term 2 using integration by parts as shown below. Consider u = x and v = e − 4 x . ∫ 0 ∞ x e − 4 x 2 d x = 1 2 [ x e − 4 x − 4 − e − 4 x 16 ] 0 ∞ = 1 2 [ − 4 x e − 4 x − e − 4 x 16 ] 0 ∞ = 1 2 [ ( − e − 4 x ) 4 x + 1 16 ] 0 ∞ = 1 2 [ 0 − ( − e − 4 ( 0 ) ) 4 ( 0 ) + 1 16 ] = 1 32 Similarly, simplify term 3 as follows. ∫ 0 ∞ e − 4 x 16 d x = ( 1 16 ) ( e − 4 x − 4 ) 0 ∞ = ( 1 16 ) ( 1 4 ) = 1 64 Substitute the above obtained values in the equation for variance as shown below. σ 2 = 4 [ 1 32 − 1 32 + 1 64 ] = 4 [ 1 64 ] = 1 16 Thus, the variance value is, σ 2 = 1 16 .

The mean and variance of the random variable X with probability density f ( x ) . The mean is μ = 1 4 _ and the variance is σ 2 = 1 16 _ . Formula used: Mean of X : μ = E ( X ) = ∫ − ∞ ∞ x f ( x ) Variance of X : σ 2 = E ( [ X − μ ] 2 ) = ∫ − ∞ ∞ ( x − μ ) 2 f ( x ) Integration by parts: Let u and v be two functions, then integration by parts states that ∫ u v d x = u v 1 − u ' v 2 + u ' ' v 3 − u ' ' ' v 4 + … . Calculation: It is given that the probability density function is f ( x ) = 4 e − 4 x ( x ≥ 0 ) . Calculate the mean as shown below. μ = ∫ 0 ∞ x 4 e − 4 x d x = 4 ∫ 0 ∞ x e − 4 x d x Consider u = x and v = e − 4 x and apply integration by parts to solve the above integration as follows. μ = 4 [ ( x e − 4 x − 4 ) 0 ∞ − ( e − 4 x 16 ) 0 ∞ ] = 4 [ 0 + 1 16 ] = 1 4 Thus, the value of mean is, μ = 1 4 . Now, calculate the variance as shown below. σ 2 = ∫ 0 ∞ ( x − 1 4 ) 2 4 e − 4 x d x = 4 ∫ 0 ∞ ( x − 1 4 ) 2 e − 4 x d x = 4 ∫ 0 ∞ ( x 2 − x 2 + 1 16 ) e − 4 x d x = 4 ∫ 0 ∞ ( x 2 e − 4 x − x e − 4 x 2 + e − 4 x 16 ) d x = 4 ( ∫ 0 ∞ x 2 e − 4 x ︸ term 1 − ∫ 0 ∞ x e − 4 x 2 ︸ term 2 + ∫ 0 ∞ e − 4 x 16 ︸ term 3 ) d x Simplify term 1 by using integration by parts, as follows. Consider u = x 2 and v = e − 4 x . ∫ 0 ∞ e − 4 x x 2 d x = ( x 2 e − 4 x − 4 − 2 x e − 4 x 16 + 2 e − 4 x − 64 ) 0 ∞ = ( x 2 e − 4 x − 4 − x e − 4 x 8 + e − 4 x − 32 ) 0 ∞ = [ 8 x 2 e − 4 x + 4 x e − 4 x + e − 4 x − 32 ] 0 ∞ = [ 1 − 32 e − 4 x ( 8 x 2 + 4 x + 1 ) ] 0 ∞ = − 1 32 Simplify term 2 using integration by parts as shown below. Consider u = x and v = e − 4 x . ∫ 0 ∞ x e − 4 x 2 d x = 1 2 [ x e − 4 x − 4 − e − 4 x 16 ] 0 ∞ = 1 2 [ − 4 x e − 4 x − e − 4 x 16 ] 0 ∞ = 1 2 [ ( − e − 4 x ) 4 x + 1 16 ] 0 ∞ = 1 2 [ 0 − ( − e − 4 ( 0 ) ) 4 ( 0 ) + 1 16 ] = 1 32 Similarly, simplify term 3 as follows. ∫ 0 ∞ e − 4 x 16 d x = ( 1 16 ) ( e − 4 x − 4 ) 0 ∞ = ( 1 16 ) ( 1 4 ) = 1 64 Substitute the above obtained values in the equation for variance as shown below. σ 2 = 4 [ 1 32 − 1 32 + 1 64 ] = 4 [ 1 64 ] = 1 16 Thus, the variance value is, σ 2 = 1 16 .

ADVANCED ENGINEERING MATHEMATICS

ADVANCED ENGINEERING MATHEMATICS

10th Edition

ISBN: 2819770198774

Author: Kreyszig

Publisher: WILEY CONS

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ADVANCED ENGINEERING MATHEMATICS

Ch. 24.1 - Prob. 1P Ch. 24.1 - Prob. 2P Ch. 24.1 - Prob. 3P Ch. 24.1 - Prob. 4P Ch. 24.1 - Prob. 5P Ch. 24.1 - Prob. 6P Ch. 24.1 - Prob. 7P Ch. 24.1 - Prob. 8P Ch. 24.1 - Prob. 9P Ch. 24.1 - Prob. 10P

Ch. 24.1 - Prob. 11P Ch. 24.1 - Prob. 12P Ch. 24.1 - Prob. 13P Ch. 24.1 - Prob. 14P Ch. 24.1 - Prob. 15P Ch. 24.1 - Prob. 16P Ch. 24.1 - Prob. 17P Ch. 24.1 - Prob. 18P Ch. 24.1 - Prob. 19P Ch. 24.1 - Prob. 20P Ch. 24.2 - Prob. 1P Ch. 24.2 - Prob. 2P Ch. 24.2 - Prob. 3P Ch. 24.2 - Prob. 4P Ch. 24.2 - Prob. 5P Ch. 24.2 - Prob. 6P Ch. 24.2 - Prob. 7P Ch. 24.2 - Prob. 8P Ch. 24.2 - Prob. 9P Ch. 24.2 - Prob. 10P Ch. 24.2 - Prob. 11P Ch. 24.2 - Prob. 12P Ch. 24.2 - Prob. 13P Ch. 24.2 - Prob. 14P Ch. 24.2 - Prob. 15P Ch. 24.2 - Prob. 16P Ch. 24.2 - Prob. 17P Ch. 24.2 - Prob. 18P Ch. 24.2 - Prob. 19P Ch. 24.2 - Prob. 20P Ch. 24.3 - Prob. 1P Ch. 24.3 - Prob. 2P Ch. 24.3 - Prob. 3P Ch. 24.3 - Prob. 4P Ch. 24.3 - Prob. 5P Ch. 24.3 - Prob. 6P Ch. 24.3 - Prob. 7P Ch. 24.3 - Prob. 8P Ch. 24.3 - Prob. 9P Ch. 24.3 - Prob. 10P Ch. 24.3 - Prob. 11P Ch. 24.3 - Prob. 12P Ch. 24.3 - Prob. 13P Ch. 24.3 - Prob. 14P Ch. 24.3 - Prob. 15P Ch. 24.3 - Prob. 16P Ch. 24.3 - Prob. 17P Ch. 24.3 - Prob. 18P Ch. 24.3 - Prob. 19P Ch. 24.4 - Prob. 1P Ch. 24.4 - Prob. 2P Ch. 24.4 - Prob. 3P Ch. 24.4 - Prob. 4P Ch. 24.4 - Prob. 5P Ch. 24.4 - Prob. 6P Ch. 24.4 - Prob. 7P Ch. 24.4 - Prob. 8P Ch. 24.4 - Prob. 9P Ch. 24.4 - Prob. 10P Ch. 24.4 - Prob. 11P Ch. 24.4 - Prob. 12P Ch. 24.4 - Prob. 15P Ch. 24.5 - Prob. 1P Ch. 24.5 - Prob. 2P Ch. 24.5 - Prob. 3P Ch. 24.5 - Prob. 4P Ch. 24.5 - Prob. 5P Ch. 24.5 - Prob. 6P Ch. 24.5 - Prob. 7P Ch. 24.5 - Prob. 8P Ch. 24.5 - Prob. 9P Ch. 24.5 - Prob. 10P Ch. 24.5 - Prob. 11P Ch. 24.5 - Prob. 12P Ch. 24.5 - Prob. 14P Ch. 24.5 - Prob. 15P Ch. 24.6 - Prob. 1P Ch. 24.6 - Prob. 2P Ch. 24.6 - Prob. 3P Ch. 24.6 - Prob. 4P Ch. 24.6 - Prob. 5P Ch. 24.6 - Prob. 6P Ch. 24.6 - Prob. 7P Ch. 24.6 - Prob. 8P Ch. 24.6 - Prob. 9P Ch. 24.6 - Prob. 10P Ch. 24.6 - Prob. 11P Ch. 24.6 - Prob. 12P Ch. 24.6 - Prob. 13P Ch. 24.6 - Prob. 14P Ch. 24.6 - Prob. 15P Ch. 24.6 - Prob. 16P Ch. 24.6 - Prob. 17P Ch. 24.6 - Prob. 18P Ch. 24.6 - Prob. 19P Ch. 24.7 - Prob. 1P Ch. 24.7 - Prob. 3P Ch. 24.7 - Prob. 4P Ch. 24.7 - Prob. 5P Ch. 24.7 - Prob. 6P Ch. 24.7 - Prob. 7P Ch. 24.7 - Prob. 8P Ch. 24.7 - Prob. 9P Ch. 24.7 - Prob. 10P Ch. 24.7 - Prob. 11P Ch. 24.7 - Prob. 12P Ch. 24.7 - Prob. 13P Ch. 24.7 - Prob. 14P Ch. 24.7 - Prob. 15P Ch. 24.7 - Prob. 17P Ch. 24.7 - Prob. 18P Ch. 24.8 - Prob. 1P Ch. 24.8 - Prob. 2P Ch. 24.8 - Prob. 3P Ch. 24.8 - Prob. 4P Ch. 24.8 - Prob. 5P Ch. 24.8 - Prob. 6P Ch. 24.8 - Prob. 7P Ch. 24.8 - Prob. 8P Ch. 24.8 - Prob. 9P Ch. 24.8 - Prob. 10P Ch. 24.8 - Prob. 11P Ch. 24.8 - Prob. 12P Ch. 24.8 - Prob. 13P Ch. 24.9 - Prob. 1P Ch. 24.9 - Prob. 2P Ch. 24.9 - Prob. 3P Ch. 24.9 - Prob. 4P Ch. 24.9 - Prob. 5P Ch. 24.9 - Prob. 6P Ch. 24.9 - Prob. 7P Ch. 24.9 - Prob. 8P Ch. 24.9 - Prob. 9P Ch. 24.9 - Prob. 10P Ch. 24.9 - Prob. 11P Ch. 24.9 - Prob. 12P Ch. 24.9 - Prob. 13P Ch. 24.9 - Prob. 14P Ch. 24.9 - Prob. 15P Ch. 24.9 - Prob. 16P Ch. 24.9 - Prob. 17P Ch. 24.9 - Prob. 18P Ch. 24.9 - Prob. 19P Ch. 24.9 - Prob. 20P Ch. 24 - Prob. 1RQ Ch. 24 - Prob. 2RQ Ch. 24 - Prob. 3RQ Ch. 24 - Prob. 4RQ Ch. 24 - Prob. 5RQ Ch. 24 - Prob. 6RQ Ch. 24 - Prob. 7RQ Ch. 24 - Prob. 9RQ Ch. 24 - Prob. 10RQ Ch. 24 - Prob. 11RQ Ch. 24 - Prob. 12RQ Ch. 24 - Prob. 13RQ Ch. 24 - Prob. 14RQ Ch. 24 - Prob. 15RQ Ch. 24 - Prob. 16RQ Ch. 24 - Prob. 17RQ Ch. 24 - Prob. 18RQ Ch. 24 - Prob. 19RQ Ch. 24 - Prob. 20RQ Ch. 24 - Prob. 21RQ Ch. 24 - Prob. 22RQ Ch. 24 - Prob. 23RQ Ch. 24 - Prob. 24RQ Ch. 24 - Prob. 25RQ

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