Explanation Given Information: The function f ( x ) = 3 4 x 2 ( 2 − x ) over the interval [ 0 , 2 ] . Formula used: A function f of a continuous random variable x in the interval [ a , b ] is a probability density function if it is nonnegative and continuous on the interval [ a , b ] and ∫ a b f ( x ) d x = 1 . The integration formula: ∫ x n d x = x n + 1 n + 1 + C The Fundamental theorem: ∫ a b f ( x ) d x = F ( b ) − F ( a ) Proof: Consider the provided function f ( x ) = 3 4 x 2 ( 2 − x ) over the interval [ 0 , 2 ] . Consider any value x = 1 in the interval [ 0 , 2 ] . Evaluate the function at x = 1 . So, f ( 1 ) = 3 4 ( 1 2 ) ( 2 − 1 ) = 3 4 So, the function f is continuous and nonnegative for all values on the interval [ 0 , 2 ] . Now, show that ∫ a b f ( x ) d x = 1 . Evaluate the integral ∫ 0 2 3 4 x 2 ( 2 − x ) d x . Use distributive property a ( b + c ) = a b + a c and clear the parenthesis

Calculus: An Applied Approach (MindTap Course List)

10th Edition

ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 9, Problem 22RE

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To prove: The function f(x)=34x2(2−x) is a probability density function in the interval [0,2].

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Chapter 9, Problem 21RE

Chapter 9, Problem 23RE

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Chapter 9 Solutions

Calculus: An Applied Approach (MindTap Course List)

Ch. 9.1 - Checkpoint 1 Worked-out solution available at...Ch. 9.1 - Prob. 2CP Ch. 9.1 - Prob. 3CP Ch. 9.1 - Prob. 4CP Ch. 9.1 - Prob. 5CP Ch. 9.1 - Prob. 6CP Ch. 9.1 - Prob. 1SWU Ch. 9.1 - Prob. 2SWU Ch. 9.1 - Prob. 3SWU Ch. 9.1 - Prob. 4SWU

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To prove: The function f ( x ) = 3 4 x 2 ( 2 − x ) is a probability density function in the interval [ 0 , 2 ] .

Solution Summary: The author proves that the function f is a probability density function in the interval left[0,2right].

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To prove: The function f(x)=34x2(2−x) is a probability density function in the interval [0,2].

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Chapter 9 Solutions